Prove $u(x,y,z)=f(t+r)/r+g(t-r)/r$ satisfies $u_{xx}+u_{yy}+u_{zz}=u_{tt}$. 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.
 A: First,  consider the function 
$$h(x,y,z)=\frac{f(t+r)}{r} \tag 1$$
and with $r^2=x^2+y^2+z^2\quad;\quad r\frac{\partial r}{\partial x}=x \quad;\quad \frac{\partial r}{\partial x}=\frac{x}{r}$
$$h_x=\left(\frac{f'(t+r)}{r}-\frac{f(t+r)}{r^2} \right)\frac{\partial r}{\partial x}=x\frac{f'}{r^2}-x\frac{f}{r^3} $$
In interest of space, we write $f$ instead of $f(t+r)$ and $f'$ instead of $f'(t+r)$ .
$$h_{xx}=\left(\frac{f'}{r^2}-\frac{f}{r^3}\right)+\left(x\frac{f''}{r^2}-x\frac{f'}{r^3} \right)\frac{x}{r}+\left(-2x\frac{f'}{r^3}+3x\frac{f}{r^4} \right)\frac{x}{r}$$
$$h_{xx}=\frac{f'}{r^2}-\frac{f}{r^3}+x^2\frac{f''}{r^3}-x^2\frac{f'}{r^4} -2x^2\frac{f'}{r^4}+3x^2\frac{f}{r^5}$$
$$h_{xx}=x^2\left(\frac{f''}{r^3}  -3\frac{f'}{r^4}+3\frac{f}{r^5}\right)+\frac{f'}{r^2}-\frac{f}{r^3}$$
On the same manner, we get :
$$h_{yy}=y^2\left(\frac{f''}{r^3}  -3\frac{f'}{r^4}+3\frac{f}{r^5}\right)+\frac{f'}{r^2}-\frac{f}{r^3}$$
$$h_{zz}=z^2\left(\frac{f''}{r^3}  -3\frac{f'}{r^4}+3\frac{f}{r^5}\right)+\frac{f'}{r^2}-\frac{f}{r^3}$$
Adding the three equations
$$h_{xx}+h_{yy}+h_{zz}=(x^2+y^2+z^2)\left(\frac{f''}{r^3}  -3\frac{f'}{r^4}+3\frac{f}{r^5}\right)+3\frac{f'}{r^2}-3\frac{f}{r^3}$$
$$h_{xx}+h_{yy}+h_{zz}=r^2\left(\frac{f''}{r^3}  -3\frac{f'}{r^4}+3\frac{f}{r^5}\right)+3\frac{f'}{r^2}-3\frac{f}{r^3}$$
$$h_{xx}+h_{yy}+h_{zz}=\frac{f''}{r}$$
From Eq.$(1)\quad:\quad h_t=\frac{f'}{r}\quad$and$\quad h_{tt}=\frac{f''}{r}\quad$thus
$$h_{xx}+h_{yy}+h_{zz}=h_{tt}$$
This proves that 
$$\frac{f(t+r)}{r}\quad \text{is solution of}\quad u_{xx}+u_{yy}+u_{zz}=u_{tt}$$
Second, on the same manner one prouves that 
$$\frac{g(t-r)}{r}\quad \text{is solution of}\quad u_{xx}+u_{yy}+u_{zz}=u_{tt}$$
Any linear combination of solutions of the PDE $(1)$ is solution of the PDE because the PDE is linear. Thus 
$$\frac{f(t+r)}{r}+\frac{g(t-r)}{r} \quad\text{satisfies}\quad u_{xx}+u_{yy}+u_{zz}=u_{tt}$$
A: Ok, so thanks to @Klaramun and @TedShifrin I managed to find the proper way to solve it. With the expression for $u_{xx}, \, u_{yy}, \, u_{zz}$ as added in the question, and noting that $r^2=x^2+y^2+z^2$, it's immediate that $$u_{xx}+u_{yy}+u_{zz} = \dfrac{1}{r}(f''+g'') = u_{tt}.$$
Hope this helps somebody :)
