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