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:
$f_x=\dfrac{y}{x^2+y^2} \\f_y=-\dfrac{x}{x^2+y^2}$
Now, $f_{xx}=\dfrac{(x^2+y^2)(0)-y \times (2x)}{(x^2+y^2)^2} =\dfrac{-2xy}{(x^2+y^2)^2} \\ f_{yy}=\dfrac{(x^2+y^2)(1)-y \times (2y)}{(x^2+y^2)^2} =\dfrac{2xy}{(x^2+y^2)^2} $
Now, $\nabla^2 f =-\dfrac{2xy}{(x^2+y^2)^2}+\dfrac{2xy}{(x^2+y^2)^2}=0$
So, the Laplace's equation is satisfied.