Answer
As a result $f_x=\frac{ady-cby}{(cx+dy)^2}$, $f_y=\frac{bcx-dax}{(cx+dy)^2}$.
Work Step by Step
$f(x,y)=\frac{ax+by}{cx+dy}$
In order to find $f_x$ we treat $y$ as a constant and differentiate with respect to $x$.
$f_x=\frac{a(cx+dy)-c(ax+by)}{(cx+dy)^2}=\frac{ady-cby}{(cx+dy)^2}$
In order to find $f_y$ we treat $x$ as a constant and differentiate with respect to $y$.
$f_y=\frac{b(cx+dy)-d(ax+by)}{(cx+dy)^2}=\frac{bcx-dax}{(cx+dy)^2}$
