Answer
$${f_x}\left( {x,y} \right) = {e^{x + y}},\,\,\,\,{f_y}\left( {x,y} \right) = {e^{x + y}},\,\,\,\,\,{f_x}\left( {2, - 1} \right) = e{\text{ and }}{f_y}\left( { - 4,3} \right) = {e^{ - 1}}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {e^{x + y}} \cr
& {\text{Find }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{e^{x + y}}} \right] \cr
& {\text{treat y as a constant and }}x{\text{ as a variable use }}\frac{d}{{dx}}\left[ {{e^u}} \right] = {e^u}\frac{{du}}{{dx}}{\text{. then}} \cr
& {f_x}\left( {x,y} \right) = {e^{x + y}}\frac{\partial }{{\partial x}}\left[ {x + y} \right] \cr
& {f_x}\left( {x,y} \right) = {e^{x + y}}\left( 1 \right) \cr
& {f_x}\left( {x,y} \right) = {e^{x + y}} \cr
& {\text{evaluate }}{f_x}\left( {2, - 1} \right) \cr
& {f_x}\left( {2, - 1} \right) = {e^{2 - 1}} \cr
& {f_x}\left( {2, - 1} \right) = e \cr
& \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{e^{x + y}}} \right] \cr
& {\text{treat x as a constant and y as a variable}}{\text{. then}} \cr
& {f_y}\left( {x,y} \right) = {e^{x + y}}\frac{\partial }{{\partial y}}\left[ {x + y} \right] \cr
& {f_y}\left( {x,y} \right) = {e^{x + y}}\left( 1 \right) \cr
& {f_y}\left( {x,y} \right) = {e^{x + y}} \cr
& {\text{evaluate }}{f_y}\left( { - 4,3} \right) \cr
& {f_y}\left( { - 4,3} \right) = {e^{ - 4 + 3}} \cr
& {f_y}\left( { - 4,3} \right) = {e^{ - 1}} \cr} $$
