If $$w=z\tan^{-1}\Big(\frac{x}{y}\Big),$$
find$$\frac{\partial^2 {w}}{\partial{x^2}}+\frac{\partial^2 {w}}{\partial{y^2}}+\frac{\partial^2 {w}}{\partial{z^2}}=?$$
I calculated this and the answer is zero. But It was long calculations ( find $\frac{\partial {w}}{\partial{x}}$ then calculate $\frac{\partial^2 {w}}{\partial{x^2}}$ and do this again for $y$ and $z$ ): \begin{align} \frac{\partial {w}}{\partial{x}}&=\frac{yz}{x^2+y^2}\\ \frac{\partial^2 {w}}{\partial{x^2}}&=\frac{2xyz}{(x^2+y^2)^2}\\ \frac{\partial {w}}{\partial{y}}&=\frac{-zx}{x^2+y^2}\\ \frac{\partial^2 {w}}{\partial{y^2}}&=\frac{-2xyz}{(x^2+y^2)^2}\\ \frac{\partial^2 {w}}{\partial{z^2}}&=0. \end{align}
I wonder if there is a shorter answer or another approach to calculate this? (Because the answer is zero I guess maybe there is different approach too.)
The second derivative with respect to $z$ is clearly zero. Furthermore, $$\arctan(y/x) = \operatorname{Im}(\operatorname{Log}(x + \mathrm iy))$$ and we know that the imaginary part of an analytic function is harmonic (satisfies Laplace's equation). The result for $\arctan(x/y)$ follows by symmetry.