Answer
$$\eqalign{
& {f_x}\left( {x,y} \right) = \frac{{{y^2}}}{{{{\left( {x + y} \right)}^2}}} \cr
& {f_y}\left( {x,y} \right) = \frac{{{x^2}}}{{{{\left( {x + y} \right)}^2}}} \cr} $$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \frac{{xy}}{{x + y}} \cr
& {\text{Calculate }}{f_x}\left( {x,y} \right){\text{ treating }}y{\text{ as a constant}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\frac{{xy}}{{x + y}}} \right] \cr
& {\text{By the quotient rule}} \cr
& {f_x}\left( {x,y} \right) = \frac{{\left( {x + y} \right)\left( y \right) - xy\left( 1 \right)}}{{{{\left( {x + y} \right)}^2}}} \cr
& {f_x}\left( {x,y} \right) = \frac{{xy + {y^2} - xy}}{{{{\left( {x + y} \right)}^2}}} \cr
& {f_x}\left( {x,y} \right) = \frac{{{y^2}}}{{{{\left( {x + y} \right)}^2}}} \cr
& \cr
& {\text{Calculate }}{f_y}\left( {x,y} \right){\text{ treating }}y{\text{ as a constant}} \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\frac{{xy}}{{x + y}}} \right] \cr
& {\text{By the quotient rule}} \cr
& {f_y}\left( {x,y} \right) = \frac{{\left( {x + y} \right)\left( x \right) - xy\left( 1 \right)}}{{{{\left( {x + y} \right)}^2}}} \cr
& {f_y}\left( {x,y} \right) = \frac{{{x^2} + xy - xy}}{{{{\left( {x + y} \right)}^2}}} \cr
& {f_y}\left( {x,y} \right) = \frac{{{x^2}}}{{{{\left( {x + y} \right)}^2}}} \cr} $$
