Answer
$\dfrac{-y}{(x^2+y^2)}$ and $\dfrac{x}{(x^2+y^2)}$
Work Step by Step
We need to take the first partial derivatives of the given function.
In order to find the partial derivative, we will differentiate with respect to $x$, by keeping $y$ as a constant, and vice versa:
$f_x=\dfrac{1}{1+(y/x)^2} \times (-x^{-2} y)=\dfrac{-y}{(x^2+y^2)}$
$f_y=\dfrac{1}{1+(y^2/x^2)^2} \times (x^{-1} )=\dfrac{x}{(x^2+y^2)}$
