Show that the function $u=\ln(x^2+y^2)$ is a solution of the two dimensional Laplace equation $$\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=0.$$
I have found an answer \begin{align*} \frac{\partial u}{\partial x} &=\frac{2x}{x^2+y^2}\\ \frac{\partial^2 u}{\partial x^2}&=\frac {2(x^2+y^2)-4x^2}{(x^2+y^2)^2} \end{align*} and $$\frac{\partial^2 u}{\partial y^2}=\frac {2(x^2+y^2)-4y^2}{(x^2+y^2)^2}$$ and added the equations and got $0$.