Answer
$f_{x}(x, y)=-\displaystyle \frac{y}{x^{2}+y^{2}},$
$f_{x}(2,3)=-\displaystyle \frac{3}{13}$.
Work Step by Step
$ f(x, y)=\arctan (y/x) =\arctan (yx^{-1})$
Treat y as constant to calculate $f_{x}(x, y)$. We need the chain rule.
$f_{x}(x, y)=\displaystyle \frac{1}{1+(yx^{-1})^{2}}\cdot\frac{\partial}{\partial x}[yx^{-1}]$
$=\displaystyle \frac{1}{1+(\frac{y}{x})^{2}}(-yx^{-2})$
$=\displaystyle \frac{-y}{x^{2}(\frac{x^{2}+y^{2}}{x^{2}})}$
$=-\displaystyle \frac{y}{x^{2}+y^{2}}$,
$f_{x}(2,3)=-\displaystyle \frac{3}{2^{2}+3^{2}}=-\frac{3}{13}$.
