Answer
$${f_x}\left( {x,y} \right) = 3{\text{ and }}{f_y}\left( {x,y} \right) = 2$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 3x + 2y \cr
& {\text{Differentiate by using the limit definition}} \cr
& {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x + h,y} \right) - f\left( {x,y} \right)}}{h} \cr
& {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{3\left( {x + h} \right) + 2y - \left( {3x + 2y} \right)}}{h} \cr
& {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{3x + 3h + 2y - 3x - 2y}}{h} \cr
& {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{3h}}{h} \cr
& {f_x}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \left( 3 \right) \cr
& {\text{Evaluate the limit}} \cr
& {f_x}\left( {x,y} \right) = 3 \cr
& \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{f\left( {x,y + h} \right) - f\left( {x,y} \right)}}{h} \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{3x + 2\left( {y + h} \right) - \left( {3x + 2y} \right)}}{h} \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{3x + 2y + 2h - 3x - 2y}}{h} \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} \frac{{2h}}{h} \cr
& {f_y}\left( {x,y} \right) = \mathop {\lim }\limits_{h \to 0} 2 \cr
& {\text{Evaluate the limit}} \cr
& {f_y}\left( {x,y} \right) = 2 \cr} $$
