Answer
The function $u=1/\sqrt{x^2+y^2+z^2}$ is a solution of the equation $u_{xx}+u_{yy}+u_{zz}=0$.
Work Step by Step
Here, we have $u_x=\dfrac{-x}{(x^2+y^2+z^2)^{3/2}}$
and $u_{xx}=\dfrac{2x^2-y^2-z^2}{(x^2+y^2+z^2)^{5/2}}$ ...(a)
The second derivative of $u$ with respect to $y$ gives:
$u_{yy}=\dfrac{2y^2-x^2-z^2}{(x^2+y^2+z^2)^{5/2}}$ ...(b)
The second derivative of $u$ with respect to $z$ gives:
$u_{zz}=\dfrac{2z^2-y^2-x^2}{(x^2+y^2+z^2)^{5/2}}$ ...(c)
We need to solve the above three equations (a), (b) and (c).
$u_{xx}+u_{yy}+u_{zz}=\dfrac{(2x^2-y^2-z^2)+(2y^2-x^2-z^2)+(2z^2-y^2-x^2)}{(x^2+y^2+z^2)^{5/2}}=0$
