I'm trying to prove that any function $u=u(x,y,z)$ of the form $$u(x,y,z)=\frac{f(t+r)}{r}+\frac{g(t-r)}{r},$$ with $r^2=x^2+y^2+z^2$, satisfies the differential equation $ u\_{xx}+u\_{yy}+u\_{zz}=u\_{tt}, $ but I'm stuck when relating the derivatives. 

From what I get, $$u\_{tt} = \dfrac{1}{r}(f''+g''),$$ as $r$ does not depend on $t$.

Furthermore, $$ u\_x = \dfrac{r\_x}{r^2}\left( r(f'-g')-(f+g) \right), $$ so, if I'm not mistaken, 

$$u\_{xx} = \dfrac{x^2}{r^3} (f''+g'') + \dfrac{r^2-3x^2}{r^4}(f'-g') + \dfrac{x^2-r^2}{r^5}(f+g). $$ 

(By symmetry, $u\_{yy}$ and $u\_{zz}$ have the same form).

But I don't know how to relate this expression to the other second derivatives, so I think I'm missing something...

Any ideas are greatly appreciated.
Thanks in advance!

EDIT: Corrected a $r^2$ term in the expression for $u\_x$. Also, thanks to @Klaramun's and @TedShifrin's suggestions, I managed to reduce the expression for $u\_{xx}$, noting that $f\_x=f\_y=f\_z=f',\, g\_x=g\_y=g\_z=g'$, so I fixed that a couple lines up.