Answer
The Laplace's equation is satisfied.
Work Step by Step
We need to compute the Laplace's equation and prove that it equals to $0$.
In order to find the partial derivative, we will differentiate with respect to $x$, by keeping $y$ and $z$ as a constant, and vice versa:
$f_x=-x (x^2+y^2+z^2)^{-3/2} \implies f_{xx}=-x (\dfrac{-3}{2}) (x^2+y^2+z^2)^{-5/2} (2x)+(x^2+y^2+z^2)^{-3/2}(-1)=(x^2+y^2+z^2)^{-5/2} (2x^2-y^2-z^2)$
and $f_y=-y (x^2+y^2+z^2)^{-3/2} \implies f_{yy}=-y (\dfrac{-3}{2}) (x^2+y^2+z^2)^{-5/2} (2y)+(x^2+y^2+z^2)^{-3/2}(-1)=(x^2+y^2+z^2)^{-5/2} (2y^2-x^2-z^2)$
and $f_z=-z (x^2+y^2+z^2)^{-3/2} \implies f_{zz}=-z (\dfrac{-3}{2}) (x^2+y^2+z^2)^{-5/2} (2z)+(x^2+y^2+z^2)^{-3/2}(-1)=(x^2+y^2+z^2)^{-5/2} (2z^2-x^2-y^2)$
Consider the Laplace's equation $\nabla^2 f =\dfrac{\partial^2 f}{\partial x^2}+\dfrac{\partial^2 f}{\partial y^2}$
$\nabla^2 f =(x^2+y^2+z^2)^{-5/2} (2x^2-y^2-z^2+2y^2-x^2-z^2+2z^2-x^2-y^2)=0$
Thus, the Laplace's equation is satisfied.
