I was also interested in answering this particular problem.
Show that $$\frac{\partial^2z}{\partial x^2} + \frac{\partial^2z}{\partial y^2}= 0$$
given the function $z=ln(x^2+y^2)$.
Now I got an crappy answer where $\frac{\partial^2z}{\partial x^2}=\frac{-x^2+3y^2+^2y^2-4xy^2-y^4}{{(x^2+y^2)}^2}$ and $\frac{\partial^2z}{\partial y^2} = \frac{x^2y^2+2x^2-4x^2y-2y^2}{{(x^2+y^2)}^2}$, which when added, doesn't equate to zero.
What's the proper way to verify the above statement?
Let me try $$\frac{\partial z}{\partial x} = \frac{2x}{x^2+y^2}.$$
$$\frac{\partial^2 z}{\partial x^2} = \frac{2(x^2+y^2) - 4x^2}{(x^2+y^2)^2} = \frac{2(y^2-x^2)}{(x^2+y^2)^2}.$$
Similarly, you have $$\frac{\partial^2 z}{\partial y^2} = \frac{2(x^2-y^2)}{(x^2+y^2)^2}.$$