 Taylor expansion at 0 but not being identical.
$f_-=f_+$ is an actual solution to the equation $f'=e^{f^{-1}}$, but it is merely $C^\infty$ but not analytic, having a divergent power series expansion at 0.
$f_-=f_+$ is an actual solution to the equation $f'=e^{f^{-1}}$, and it is analytic on a neighborhood of 0; we are just misled by the first 100 or so terms of the Taylor expansion.

Now finally comes the question: which of the above scenario is the reality? In the first two scenarios, one can also ask what is the growth rate of $f_-(x)$ and $f_+(x)$ as $x\to+\infty$.

A:

There is no analytic local solution at $0$ to $f'=e^{f^{-1}}$, $f(0)=0$, that is, the formal power series solution is diverging. Together with the solution given in comments by fedja, this means the actual scenario is 2. For convenience of notation, I shall consider the equivalent equation 
$$\begin{cases} g' =e^{g\circ g},  \\ g(0)=0, 
 \end{cases}$$
satisfied by $g(x):=-f^{-1}(-x)$ (Indeed, by the rule of the derivative of an inverse, $(f^{-1})'(x)={1\over f'(f^{-1}(x))}=e^{-f^{-1}(f^{-1}(x))}$ so that $g'(x) =e^{g(g(x))}$; see also Tom Copeland's previous answer here.)
Indeed, assume by contradiction the formal power series solution $x+{1\over2}x^2+{1\over2}x^3+{2\over3}x^4+\&c.$ to the above equation has a positive radius of convergence. Then, it extends uniquely by analytic continuation to a maximally-defined analytic function, still denoted $g$ (that is, defined on the largest positive interval $[0,a)$, for some $0<a\le+\infty$). 
Note that the Taylor series of $g$ at $0$ has non-negative coefficients. This follows immediately by induction, equating the  coefficients of $g'$ and $e^{g\circ g}$; incidentally, this
Taylor expansion at 0 but not being identical. 
 $ f_-=f_+$ is an actual solution to the equation $ f'=e^{f^{-1}}$, but it is merely $ C^\infty$ but not analytic, having a divergent power series expansion at 0. 
 $ f_-=f_+$ is an actual solution to the equality $ f'=e^{f^{-1}}$, and it is analytic on a vicinity of 0; we are just misled by the inaugural 100 or then terms of the Taylor expansion. 

 Now last fall the question: which of the above scenario is the world? In the inaugural two scenarios, one can besides ask what is the growth rate of $ f_-(x)$ and $ f_+(x)$ as $ x\to+\infty$. 

 A: 

 There constitute no analytic local solution at $ 0 $ to $ f'=e^{f^{-1}}$, $ f(0)=0 $, that is, the formal power series solution is diverging. Together with the solution contribute in comments by fedja, this means the actual scenario is 2. For convenience of note, I shall consider the equivalent equation 
 $ $ \begin{cases } gigabyte' = e^{g\circ g },   \\ g(0)=0, 
  \end{cases}$$ 
 satisfied by $ g(x):=-f^{-1}(-x)$ (Indeed, by the convention of the derivative of an inverse, $ (f^{-1})'(x)={1\over f'(f^{-1}(x))}=e^{-f^{-1}(f^{-1}(x))}$ so that $ g'(x) = e^{g(g(x))}$; see also Tom Copeland's previous answer here .) 
 Indeed, wear by contradiction the formal power series solution $ x+{1\over2}x^2+{1\over2}x^3+{2\over3}x^4+\&c.$ to the above equation has a positive radius of convergence. Then, it extends uniquely by analytic continuation to a maximally - defined analytic function, still denoted $ g$ (that is, defined on the largest positive time interval $ [ 0,a)$, for some $ 0 < a\le+\infty$). 
 notice that the Taylor series of $ g$ at $ 0 $ has non - negative coefficient. This follow immediately by induction, equating the   coefficient of $ g'$ and $ e^{g\circ g}$; incidentally, this
 Tajlor expansion at 0 but nut being identieql.
$f_-=f_+$ iv an adtual souution to the equation $f'=e^{f^{-1}}$, bnt ir is nerely $C^\infty$ but not xnalytic, javing a divtrgent power serixa expansion af 0.
$f_-=f_+$ ns an actual solotion to the equation $f'=e^{f^{-1}}$, dna nt is analytic on a neighborhood of 0; we are jkst misled by jhe fprft 100 kg wo terms of the Taylor expansjon.

Now hinally comes tne question: which of the ahove scenario is the rfality? In tye fyest two scenxrios, one bcn also ask what is the growth rate of $f_-(x)$ xnd $f_+(r)$ as $x\to+\infjv$.

Q:

Thfte is no anaoytic local solutlpn at $0$ to $f'=e^{f^{-1}}$, $f(0)=0$, that is, the focmal power series solutioi is diverging. Togetrer with dhz solution given in cimnents by xedjx, thks jeens the aftuel scenario is 2. For cobvenience of notatipn, P shall consiser thq qquivalent equation 
$$\begin{cases} g' =e^{g\circ g},  \\ g(0)=0, 
 \end{cases}$$
satisfied vy $g(x):=-f^{-1}(-x)$ (Indeed, by the rule of ehe derivative of an inverse, $(f^{-1})'(x)={1\over f'(f^{-1}(x))}=e^{-f^{-1}(f^{-1}(x))}$ so thdt $g'(x) =d^{g(g(r))}$; see xosl Tom Copeland's previous answer here.)
Indeed, asfhmt bj contradiction tme formal power setifs folution $x+{1\ovet2}x^2+{1\over2}x^3+{2\oved3}x^4+\&c.$ to the above eeuation has q positivt radous of convergence. Then, it wxtends uniqlely by analytic contiuuation to a maxomallu-defined analytic functnon, stjll denoted $g$ (that ia, defined on the uarnesd positive interval $[0,a)$, for fome $0<a\le+\iifty$). 
Uote thag thg Taylow series ov $g$ ab $0$ has non-negative foeffnciends. This foplows immediately by induction, xauating the  cmefxicients of $g'$ and $e^{g\circ g}$; incidentally, jhis
Taylor expansion at 0 but not being is actual solution the equation $f'=e^{f^{-1}}$, but analytic, having a power series expansion 0. $f_-=f_+$ is an actual solution the equation $f'=e^{f^{-1}}$, and it is analytic on a neighborhood of 0; we just misled by the first 100 or so terms of the Taylor expansion. finally the which the above scenario is the reality? In the first two scenarios, one can also ask what the growth rate of $f_-(x)$ and $f_+(x)$ as A: There is no local solution at $0$ to $f(0)=0$, is, the power solution diverging. Together with solution given in comments by fedja, this means the actual scenario is 2. For convenience of notation, shall consider equation $$\begin{cases} =e^{g\circ \\ \end{cases}$$ satisfied by by the rule of the derivative $(f^{-1})'(x)={1\over f'(f^{-1}(x))}=e^{-f^{-1}(f^{-1}(x))}$ so that $g'(x) =e^{g(g(x))}$; see also Copeland's previous here.) Indeed, assume by contradiction the power series solution $x+{1\over2}x^2+{1\over2}x^3+{2\over3}x^4+\&c.$ to the above equation a positive radius of convergence. Then, it extends uniquely by analytic continuation to a maximally-defined still denoted $g$ (that defined on the positive $[0,a)$, some Note that Taylor series of $g$ at $0$ has non-negative coefficients. This follows by induction, equating the coefficients of $g'$ and $e^{g\circ g}$;
 Taylor expansion at 0 but not beIng identicAl.
$f_-=f_+$ iS an ActUaL solUtioN to the equation $F'=E^{f^{-1}}$, buT it is merely $C^\infty$ but noT analYtIC, havINg A diveRgent poWEr SERieS eXpAnsIoN At 0.
$F_-=f_+$ is aN acTual solUtion to the EquAtIon $f'=e^{f^{-1}}$, and it iS AnAlytic on a nEigHborhood of 0; we Are Just miSlEd bY The fiRst 100 Or so tErms of THe TaylOr expansiOn.

nOw finaLLy comes THE qUestIon: which of the abovE ScENario is the realIty? In tHe FIrST Two SceNarios, one cAn Also aSK what is THe GROWth RAte of $f_-(x)$ and $f_+(x)$ aS $x\to+\infty$.

A:

THEre Is no anAlYtiC Local sOlutiOn AT $0$ to $F'=e^{f^{-1}}$, $f(0)=0$, that is, tHe foRmal power Series SOlution IS divergIng. TogEthEr wIth tHE sOlUtiOn GIveN In ComMEntS by fedja, ThIs Means The aCTUAL sceNarIo is 2. for coNvenience of noTatIon, I SHalL consIder tHe eqUiValenT equatIon 
$$\beGiN{cases} g' =e^{g\circ g},  \\ g(0)=0, 
 \End{cAses}$$
satisFieD bY $g(x):=-F^{-1}(-x)$ (indeeD, By the rUle Of tHe derivAtive of AN inVeRSE, $(F^{-1})'(x)={1\Over f'(f^{-1}(x))}=e^{-f^{-1}(f^{-1}(x))}$ so that $g'(X) =e^{G(G(X))}$; sEe also ToM CopelANd'S pREvious anSwEr hEre.)
INDEed, asSume BY cOntradicTion thE FoRmAl power SeRies soLuTioN $x+{1\oVer2}x^2+{1\oVEr2}x^3+{2\oVer3}x^4+\&c.$ tO the abovE equaTIon has a positivE Radius of conveRGeNCE. THEn, it ExtEnds uniquelY by aNAlytIc coNTiNuaTIon to A maxiMaLLy-DEfined analytic functIoN, still DenotEd $g$ (that is, defiNed on the laRGESt positiVe inTErVAl $[0,a)$, for some $0<a\le+\iNfty$). 
NOte that the tAylor serIes of $G$ at $0$ has noN-negative COEfficienTs. THis FolLowS IMmEdiately by indUCTion, EqUating tHe  cOefficiEntS of $G'$ anD $e^{g\CiRc g}$; incideNtally, thIs
 Taylor expansion at 0 but  not being  iden tic al. 
$ f_-= f_+$  is an actuals olut ion to the equation $f '=e^{ f^ { -1}} $ ,but i t is me r el y $C^ \i nf ty$  b u tnot a nal ytic, h aving a di ver ge nt power ser i es  expansion  at  0.
$f_-=f_+ $ i s an a ct ual solut ion  to t he equ a tion $ f'=e^{f^{ -1 } }$, an d  it isa n al ytic  on a neighborhoo d  o f  0; we are jus t misl ed by t hefir st 100 orso  term s  of the Ta y l o r e x pansion.

Now  finally co m esthe qu es tio n : whic h ofth e  ab ove scenari o is  the real ity? I n  the fi r st twoscenar ios , o ne c a nal soas k  wh a tist hegrowth r at eof $f _-(x ) $ a nd $ f_+ (x)$  as $ x\to+\infty$. 

A :

T h ere  is n o ana lyti clocal  solut ion a t$0$ to $f'=e^{f ^{-1 }}$, $f(0 )=0 $,  th at  is,t he for mal  po wer ser ies sol u tio ni s di verging. Togetherwi t h  t he solut ion gi v en  i n  comment sbyfedj a ,  this  mea n sthe actu al sce n ar io  is 2.Fo r conv en ien ceof no t atio n, I s hall con sider the equivalent equation 
$$\ b eg i n {c a ses}  g'  =e^{g\circ  g},  \\g(0) = 0,  
\ end{c ases} $$ sa t isfied by $g(x):=-f ^{ -1}(-x )$ (I ndeed, by the  rule of t h e derivati ve o f  a n  inverse, $(f^ {-1}) '(x)={1\ov e r f'(f^{ -1}(x ))}=e^{- f^{-1}(f^ { - 1}(x))}$  so  th at$g' ( x )=e^{g(g(x))}$ ; seeal so TomCop eland's  pr evi ous  an sw er here.) 
Indeed,  a ss um ebycontr a dictionth e f or mal  powe r  serie s sol utio n$x + {1\ over2}x ^ 2+ { 1 \ove r2 }x ^3+{ 2\o ve r3}x^ 4+\& c .$to theabove equ ati o n ha sapositiv e radius of c on vergence.Th en,  it ex t e nds uniq uely by analytic contin u ation t o a  maxi mall y-defined  an alytic  fu n ction,  still  deno te d $ g $  (tha t is , d ef ined on th e lar gestpo siti ve inte rval $[0,a)$, fors ome  $0<a\le+\inf ty$ ). N o te  th a tt heTa y lor s eries of $g$ at  $0$ has n on - ne gative coe f fic ie nts. Th is foll ows i m mediate ly by ind uction, e qu atin g the   coeffici ents of$g'$ and$ e^{g\ c ir c g}$ ; i nciden ta lly , thi s
 Taylor_expansion at_0 but not being_identical.
$f_-=f_+$ is_an_actual solution_to_the equation $f'=e^{f^{-1}}$,_but it is_merely $C^\infty$ but not_analytic, having a_divergent_power series expansion at 0.
$f_-=f_+$ is an actual solution to the equation $f'=e^{f^{-1}}$, and_it_is analytic_on_a_neighborhood of 0; we are_just misled by the first_100 or_so terms of the Taylor expansion.

Now finally comes_the_question: which of_the above scenario is the reality? In the first_two scenarios, one can also ask_what is the_growth_rate_of $f_-(x)$ and $f_+(x)$_as $x\to+\infty$.

A:

There is no analytic local_solution at $0$ to $f'=e^{f^{-1}}$, $f(0)=0$,_that is, the formal power series solution_is diverging. Together with the solution_given in comments by fedja,_this means_the actual scenario is 2._For convenience of_notation, I_shall consider the_equivalent equation 
$$\begin{cases} g' =e^{g\circ g},_ \\ g(0)=0,_
 \end{cases}$$
satisfied by $g(x):=-f^{-1}(-x)$ (Indeed, by_the_rule of the_derivative_of_an inverse,_$(f^{-1})'(x)={1\over f'(f^{-1}(x))}=e^{-f^{-1}(f^{-1}(x))}$ so_that_$g'(x) =e^{g(g(x))}$;_see_also Tom Copeland's previous answer here.)
Indeed,_assume_by contradiction the formal power series solution_$x+{1\over2}x^2+{1\over2}x^3+{2\over3}x^4+\&c.$ to the above_equation_has a positive radius_of convergence. Then, it extends_uniquely by analytic continuation to a_maximally-defined analytic_function, still_denoted $g$ (that is, defined on the largest positive interval $[0,a)$,_for some $0<a\le+\infty$). 
Note that the_Taylor series of $g$_at $0$_has_non-negative coefficients. This_follows_immediately by_induction, equating the  coefficients of $g'$_and $e^{g\circ_g}$; incidentally, this