Answer
$- \dfrac{1}{(x+y)^{2} }$
and
$- \dfrac{1}{(x+y)^{2} }$
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 $y$ as a constant, and vice versa:
$f_x=-(x+y)^{-2} =- \dfrac{1}{(x+y)^{2} }$
and
$f_y=-(x+y)^{-2} =- \dfrac{1}{(x+y)^{2} }$
