Let $f(x,y,z) = g(\sqrt{x^2 + y^2 +z^2})$, where $g(r)$ is twice differentiable. Calculate $f_{xx}$ + $f_{yy}$ + $f_{zz}$ in terms of the derivatives of $g$.
Prove that if $f_{xx}$ + $f_{yy}$ + $f_{zz}$ $= 0$, then $g(r) = a/r + b$, where $a$ and $b$ are constants.
I've been able to do the first part of the question and I got that $f_{xx}$ + $f_{yy}$ + $f_{zz}$ $=$ $\frac{g''(r)(x + y + z) + 3g'(r) - 1}{\sqrt{x^2 + y^2 + z^2}}$
But I cannot figure out how to prove that $g(r) = a/r + b$. I've tried everything and can't seem to be able to show it.
Can some one please give me a hand? Any assistance would be really appreciated as I have been working on this question for a few days.