Answer
Laplace's equation is satisfied.
Work Step by Step
We need to verify Laplace's equation
$\nabla^2 f =\dfrac{\partial^2 f}{\partial x^2}+\dfrac{\partial^2 f}{\partial y^2}=0$
We take the first partial derivative of the given function $f(x,y)$ with respect to $x$, by treating $y$ as a constant, and vice versa:
$f_x=\dfrac{x}{x^2+y^2} \\f_y=\dfrac{y}{x^2+y^2}$
and $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} \\ 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} $
Now, $\nabla^2 f =\dfrac{y^2-x^2}{(x^2+y^2)^2}+\dfrac{x^2-y^2}{(x^2+y^2)^2}=0$
It has been proved that the Laplace's equation is satisfied.