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, $u_x=\dfrac{-x}{(x^2+y^2+z^2)^{3/2}}$
Now, take the second derivative of $u$ with respect to $x$.
$u_{xx}=\dfrac{2x^2-y^2-z^2}{(x^2+y^2+z^2)^{5/2}}$ ...(1)
Now, take the second derivative of $u$ with respect to $y$.
$u_{yy}=\dfrac{2y^2-x^2-z^2}{(x^2+y^2+z^2)^{5/2}}$ ...(2)
Now, take the second derivative of $u$ with respect to $z$.
$u_{zz}=\dfrac{2z^2-y^2-x^2}{(x^2+y^2+z^2)^{5/2}}$ ...(3)
From equations (1), (2) and (3) we have
$u_{xx}+u_{yy}+u_{zz}=\dfrac{(2x^2-y^2-z^2)+(2y^2-x^2-z^2)+(2z^2-y^2-x^2)}{(x^2+y^2+z^2)^{5/2}}=0$
Hence, it has been proved that the function $u=1/\sqrt{x^2+y^2+z^2}$ is a solution of the equation $u_{xx}+u_{yy}+u_{zz}=0$.