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