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