Answer
$${f_x}\left( {x,y} \right) = - \frac{1}{{{{\left( {x + y} \right)}^2}}}{\text{ and }}{f_y}\left( {x,y} \right) = - \frac{1}{{{{\left( {x + y} \right)}^2}}}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \frac{1}{{x + y}} \cr
& {\text{Differentiate by using the limit definition}} \cr
& {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h,y} \right) - f\left( {x,y} \right)}}{h} \cr
& {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{{x + h + y}} - \frac{1}{{x + y}}}}{h} \cr
& {\text{Rationalizing}} \cr
& {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{x + y - x - h - y}}{{\left( {x + h + y} \right)\left( {x + y} \right)}}}}{h} \cr
& {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{ - h}}{{\left( {x + h + y} \right)\left( {x + y} \right)}}}}{h} \cr
& {f_x}\left( {x,y} \right) = - \mathop {\lim }\limits_{h \to 0} \frac{1}{{\left( {x + h + y} \right)\left( {x + y} \right)}} \cr
& {\text{Evaluate the limit}} \cr
& {f_x}\left( {x,y} \right) = - \frac{1}{{\left( {x + 0 + y} \right)\left( {x + y} \right)}} \cr
& {f_x}\left( {x,y} \right) = - \frac{1}{{{{\left( {x + y} \right)}^2}}} \cr
& \cr
& {\text{Differentiate by using the limit definition}} \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{1}{{x + y + h}} - \frac{1}{{x + y}}}}{h} \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{x + y - x - y - h}}{{\left( {x + y + h} \right)\left( {x + y} \right)}}}}{h} \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{\frac{{ - h}}{{\left( {x + y + h} \right)\left( {x + y} \right)}}}}{h} \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{ - h}}{{h\left( {x + y + h} \right)\left( {x + y} \right)}} \cr
& {f_y}\left( {x,y} \right) = - \mathop {\lim }\limits_{h \to 0} \frac{1}{{\left( {x + y + h} \right)\left( {x + y} \right)}} \cr
& {\text{Evaluate the limit}} \cr
& {f_y}\left( {x,y} \right) = - \frac{1}{{\left( {x + y} \right)\left( {x + y} \right)}} \cr
& {f_y}\left( {x,y} \right) = - \frac{1}{{{{\left( {x + y} \right)}^2}}} \cr} $$