Answer
$f_x=\dfrac{-y}{(x^2+y^2)}$
$f_y=\dfrac{x}{(x^2+y^2)}$
Work Step by Step
Our aim is to take the first partial derivatives of the given function $f(x,y)$ with respect to $x$, by keeping $y$ and $z$ as a constant, and vice versa:
$f_x=\dfrac{1}{1+(\dfrac{y}{x})^2} \times (-x^{-2} y)=\dfrac{-y}{(x^2+y^2)}$
$f_y=\dfrac{1}{1+(\dfrac{y^2}{x^2})} \times (x^{-1} )=\dfrac{x}{(x^2+y^2)}$
