Answer
$${f_x}\left( {x,y} \right) = - 24{e^{4x - 3y}},\,\,\,\,{f_y}\left( {x,y} \right) = 18{e^{4x - 3y}},\,\,\,\,\,{f_x}\left( {2, - 1} \right) = - 24{e^{11}}{\text{ and }}{f_y}\left( { - 4,3} \right) = 18{e^{ - 25}}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = - 6{e^{4x - 3y}} \cr
& {\text{find }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ { - 6{e^{4x - 3y}}} \right] \cr
& {\text{treat y as a constant and }}x{\text{ as a variable use }}\frac{d}{{dx}}\left[ {{e^u}} \right] = {e^u}\frac{{du}}{{dx}}{\text{. then}} \cr
& {f_x}\left( {x,y} \right) = - 6{e^{4x - 3y}}\frac{\partial }{{\partial x}}\left[ {4x - 3y} \right] \cr
& {f_x}\left( {x,y} \right) = - 6{e^{4x - 3y}}\left( 4 \right) \cr
& {f_x}\left( {x,y} \right) = - 24{e^{4x - 3y}} \cr
& {\text{evaluate }}{f_x}\left( {2, - 1} \right) \cr
& {f_x}\left( {2, - 1} \right) = - 24{e^{4\left( 2 \right) - 3\left( { - 1} \right)}} \cr
& {f_x}\left( {2, - 1} \right) = - 24{e^{11}} \cr
& \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - 6{e^{4x - 3y}}} \right] \cr
& {\text{treat x as a constant and y as a variable}}{\text{. then}} \cr
& {f_y}\left( {x,y} \right) = - 6{e^{4x - 3y}}\frac{\partial }{{\partial y}}\left[ {4x - 3y} \right] \cr
& {f_y}\left( {x,y} \right) = - 6{e^{4x - 3y}}\left( { - 3} \right) \cr
& {f_y}\left( {x,y} \right) = 18{e^{4x - 3y}} \cr
& {\text{evaluate }}{f_y}\left( { - 4,3} \right) \cr
& {f_y}\left( { - 4,3} \right) = 18{e^{4\left( { - 4} \right) - 3\left( 3 \right)}} \cr
& {f_y}\left( { - 4,3} \right) = 18{e^{ - 16 - 9}} \cr
& {f_y}\left( { - 4,3} \right) = 18{e^{ - 25}} \cr} $$
