Answer
$${f_x}\left( {x,y} \right) = 5y{\text{ and }}{f_y}\left( {x,y} \right) = 5x$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 5xy \cr
& {\text{Differentiate by using the limit definition}} \cr
& {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h,y} \right) - f\left( {x,y} \right)}}{h} \cr
& {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{5\left( {x + h} \right)y - 5xy}}{h} \cr
& {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{5xy + 5hy - 5xy}}{h} \cr
& {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{5hy}}{h} \cr
& {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} 5y \cr
& {\text{Evaluate the limit}} \cr
& {f_x}\left( {x,y} \right) = 5y \cr
& \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x,y + h} \right) - f\left( {x,y} \right)}}{h} \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{5x\left( {y + h} \right) - 5xy}}{h} \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{5xy + 5xh - 5xy}}{h} \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{5xh}}{h} \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} 5x \cr
& {\text{Evaluate the limit}} \cr
& {f_y}\left( {x,y} \right) = 5x \cr} $$
