Laplace equation of composite function

The simplest method is to change the Cartesian to the polar cordinates. $$\frac{\partial^2 f}{\partial X^2}+\frac{\partial^2 f}{\partial Y^2}=0 \to \begin{cases} X=R\cos(a)\\ Y=R\sin(a)\\ \frac{\partial^2 f}{\partial R^2}+\frac{1}{R}\frac{\partial f}{\partial R}+ \frac{1}{R^2} \frac{\partial^2 f}{\partial a^2}=0 \end{cases}$$ $$\begin{cases} x=r\cos(a)\\ y=r\sin(a)\\ X=\frac{x}{x^2+y^2}=\frac{1}{r}\cos(a)\\ Y=\frac{y}{x^2+y^2}=\frac{1}{r}\sin(a)\\ \end{cases} \to r=\frac{1}{R}$$

$\frac{\partial f}{\partial R}=\frac{\partial f}{\partial r}\frac{dr}{dR}=-\frac{\partial f}{\partial r}\frac{1}{R^2}$

$\frac{\partial^2 f}{\partial R^2}=\frac{\partial^2 f}{\partial r^2}\frac{1}{R^4}+\frac{\partial f}{\partial r}\frac{2}{R^3}$

$$\frac{\partial^2 f}{\partial R^2}+\frac{1}{R}\frac{\partial f}{\partial R}+\frac{\partial^2 f}{\partial a^2}= \frac{\partial^2 f}{\partial r^2}\frac{1}{R^4}+\frac{\partial f}{\partial r}\frac{1}{R^3} +\frac{1}{R^2}\frac{\partial^2 f}{\partial a^2}=0 $$ $$ \frac{\partial^2 f}{\partial r^2}+\frac{\partial f}{\partial r}\frac{1}{r} +\frac{1}{r^2}\frac{\partial^2 f}{\partial a^2}=0 $$ Then, comming back to the cartesian coodinates : $$\left(\frac{\partial^2 }{\partial x^2}+\frac{\partial^2 }{\partial y^2}\right)f\left(\frac{x}{x^2+y^2}\:,\:\frac{y}{x^2+y^2} \right)=0$$ So, the answer to the question is yes.