Answer
$${\text{ }}{f_{xx}}\left( {x,y} \right) = 8{e^{2y}}{\text{ and }}{f_{xy}}\left( {x,y} \right) = 16x{e^{2y}}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 4{x^2}{e^{2y}} \cr
& {\text{find }}{f_x}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {4{x^2}{e^{2y}}} \right] \cr
& {\text{treat y as a constant and }}x{\text{ as a variable}}{\text{. then}} \cr
& {f_x}\left( {x,y} \right) = 4{e^{2y}}\frac{\partial }{{\partial x}}\left[ {{x^2}} \right] \cr
& {\text{solving derivatives}} \cr
& {f_x}\left( {x,y} \right) = 4{e^{2y}}\left( {2x} \right) \cr
& {f_x}\left( {x,y} \right) = 8x{e^{2y}} \cr
& \cr
& {\text{find }}{f_{xx}}\left( {x,y} \right) \cr
& {\text{ }}{f_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {8x{e^{2y}}} \right] \cr
& {\text{ }}{f_{xx}}\left( {x,y} \right) = 8{e^{2y}}\frac{\partial }{{\partial x}}\left[ x \right] \cr
& {\text{ }}{f_{xx}}\left( {x,y} \right) = 8{e^{2y}} \cr
& \cr
& {\text{find }}{f_{xy}}\left( {x,y} \right) \cr
& {\text{differentiate }}{f_x}\left( {x,y} \right){\text{ with respect to }}y \cr
& {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {8x{e^{2y}}} \right] \cr
& {\text{treat }}x{\text{ as a constant and }}y{\text{ as a variable}}{\text{. then}} \cr
& {f_{xy}}\left( {x,y} \right) = 8x\frac{\partial }{{\partial y}}\left[ {{e^{2y}}} \right] \cr
& {f_{xy}}\left( {x,y} \right) = 8x\left( {2{e^{2y}}} \right) \cr
& {f_{xy}}\left( {x,y} \right) = 16x{e^{2y}} \cr} $$
