Answer
$${f_x}\left( {x,y} \right) = \frac{{ - 2y}}{{{{\left( {x - y} \right)}^2}}},\,\,\,\,\,\,{f_y}\left( {x,y} \right) = \frac{{2x}}{{{{\left( {x - y} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \frac{{x + y}}{{x - y}} \cr
& {\text{Calculate }}{f_x}\left( {x,y} \right){\text{ differentiating with respect to }}x;{\text{ treat }}y{\text{ as a constant}} \cr
& {\text{use the quotient rule}} \cr
& {f_x}\left( {x,y} \right) = \frac{{\left( {x - y} \right)\partial /\partial x\left( {x + y} \right) - \left( {x + y} \right)\partial /\partial x\left( {x - y} \right)}}{{{{\left( {x - y} \right)}^2}}} \cr
& {f_x}\left( {x,y} \right) = \frac{{\left( {x - y} \right)\left( {1 + 0} \right) - \left( {x + y} \right)\left( {1 - 0} \right)}}{{{{\left( {x - y} \right)}^2}}} \cr
& {\text{simplifying}} \cr
& {f_x}\left( {x,y} \right) = \frac{{x - y - x - y}}{{{{\left( {x - y} \right)}^2}}} \cr
& {f_x}\left( {x,y} \right) = \frac{{ - 2y}}{{{{\left( {x - y} \right)}^2}}} \cr
& \cr
& {\text{Calculate }}{f_y}\left( {x,y} \right){\text{ differentiating with respect to }}y;{\text{ treat }}x{\text{ as a constant}} \cr
& {\text{use the quotient rule}} \cr
& {f_y}\left( {x,y} \right) = \frac{{\left( {x - y} \right)\partial /\partial y\left( {x + y} \right) - \left( {x + y} \right)\partial /\partial y\left( {x - y} \right)}}{{{{\left( {x - y} \right)}^2}}} \cr
& {f_y}\left( {x,y} \right) = \frac{{\left( {x - y} \right)\left( {0 + 1} \right) - \left( {x + y} \right)\left( {0 - 1} \right)}}{{{{\left( {x - y} \right)}^2}}} \cr
& {\text{simplifying}} \cr
& {f_y}\left( {x,y} \right) = \frac{{x - y + x + y}}{{{{\left( {x - y} \right)}^2}}} \cr
& {f_y}\left( {x,y} \right) = \frac{{2x}}{{{{\left( {x - y} \right)}^2}}} \cr} $$
