Answer
$$\frac{\partial}{\partial x}f(x,y)=\frac{(ad-bc)y}{(cx+dy)^2};$$
$$\frac{\partial}{\partial y}f(x,y)=\frac{(bc-ad)x}{(cx+dy)^2}.$$
Work Step by Step
Regard $y$ as constant and differentiate with respect to $x$:
$$\frac{\partial}{\partial x}f(x,y)=\frac{\partial}{\partial x}\left(\frac{ax+by}{cx+dy}\right)=\frac{(ax+by)'_x(cx+dy)-(ax+by)(cx+dy)'_x}{(cx+dy)^2}=\frac{a(cx+dy)-c(ax+by)}{(cx+dy)^2}=\frac{acx+ady-acx-bcy}{(cx+dy)^2}=\frac{(ad-bc)y}{(cx+dy)^2}.$$
Regard $x$ as constant and differentiate with respect to $y$:
$$\frac{\partial}{\partial y}f(x,y)=\frac{\partial}{\partial y}\left(\frac{ax+by}{cx+dy}\right)=\frac{(ax+by)'_y(cx+dy)-(ax+by)(cx+dy)'_y}{(cx+dy)^2}=\frac{b(cx+dy)-d(ax+by)}{(cx+dy)^2}=\frac{bcx+bdy-adx-bdy}{(cx+dy)^2}=\frac{(bc-ad)x}{(cx+dy)^2}.$$
