Answer
$$\eqalign{
& {f_x}\left( {x,y} \right) = 4{y^3}{e^{4x}} \cr
& {f_y}\left( {x,y} \right) = 3{y^2}{e^{4x}} \cr} $$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {y^3}{e^{4x}} \cr
& {\text{Calculate }}{f_x}\left( {x,y} \right){\text{ treating }}y{\text{ as a constant}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{y^3}{e^{4x}}} \right] \cr
& {f_x}\left( {x,y} \right) = {y^3}\frac{\partial }{{\partial x}}\left[ {{e^{4x}}} \right] \cr
& {f_x}\left( {x,y} \right) = {y^3}\left( {4{e^{4x}}} \right) \cr
& {f_x}\left( {x,y} \right) = 4{y^3}{e^{4x}} \cr
& {\text{Calculate }}{f_y}\left( {x,y} \right){\text{ treating }}x{\text{ as a constant}} \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{y^3}{e^{4x}}} \right] \cr
& {f_y}\left( {x,y} \right) = {e^{4x}}\frac{\partial }{{\partial y}}\left[ {{y^3}} \right] \cr
& {f_y}\left( {x,y} \right) = {e^{4x}}\left( {3{y^2}} \right) \cr
& {f_y}\left( {x,y} \right) = 3{y^2}{e^{4x}} \cr} $$
