partial/partial z = tan^(-1) (z/x) .
Thank you very much.
recall that $\displaystyle \frac d{dx} \arctan u = \frac {u'}{1 + u^2}$ where u is a function of x
we are taking the derivative with respect to z, so we treat x as if it was a constant.
Let $\displaystyle f(x,y,z) = \tan^{-1} \left( \frac zx \right)$
$\displaystyle \Rightarrow \frac {\partial f}{\partial z} = \frac 1{1 + \left( \frac zx \right)^2} \cdot \frac 1x$
and simplify