Small signal analysis of diode connected Field effect transistor
Since the diode connected transistor shown in your question is a NMOS (arrow marked outwards at the bottom), first analyze the small signal model of a normal NMOS. It should look something like the one shown below,

Now since this NMOS is diode connected the gate voltage is equal to the drain voltage, Vg = Vd. And since we are going to do a small signal analysis all the large signal voltages has to be shorted. This mean Vd=0V and as Vd=Vg , Vg=0V too. So the circuit would look something like this now,

You can see that both Vg and Vd are connected together(diode connected) and Vd is grounded.Rout from below can be found assuming a voltage source below that generates a current Ix and voltage of Vx. So if Vx/Ix is found that should give you the required Rout. Performing a nodal anaylsis we get the following equations,
Ix = Vx/ro + gm*Vx = Vx (1/ro + gm) (Since Vd is grounded it is at 0V and since Vg = 0V Vgs = -Vx as Vx is present at the source)
Rout=Vx/Ix = 1/(1/ro + gm) = 1/gm because ro is a big value and 1/ro will be negligible.