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=\dfrac{x}{x^2+y^2} \implies f_{xx}=\dfrac{(x^2+y^2)(1)-x(2x)}{(x^2+y^2)^2} =\dfrac{y^2-x^2}{(x^2+y^2)^2} $
and $f_y=\dfrac{y}{x^2+y^2} \implies f_{yy}=\dfrac{(x^2+y^2)(1)-y(2y)}{(x^2+y^2)^2} =\dfrac{x^2-y^2}{(x^2+y^2)^2} $
Consider the Laplace's equation $\nabla^2 f =\dfrac{\partial^2 f}{\partial x^2}+\dfrac{\partial^2 f}{\partial y^2}$
$\nabla^2 f =\dfrac{y^2-x^2}{(x^2+y^2)^2}+\dfrac{x^2-y^2}{(x^2+y^2)^2}=0$
Thus, the Laplace's equation is satisfied.
