Answer
$${f_x}\left( {x,y} \right) = 3{x^2}{e^{3y}}{\text{ and }}{f_y}\left( {x,y} \right) = 3{x^3}{e^{3y}}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {x^3}{e^{3y}} \cr
& {\text{treat y as a constant and }}x{\text{ as a variable}}{\text{. then}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{x^3}{e^{3y}}} \right] \cr
& {f_x}\left( {x,y} \right) = {e^{3y}}\frac{\partial }{{\partial x}}\left[ {{x^3}} \right] \cr
& {f_x}\left( {x,y} \right) = {e^{3y}}\left( {3{x^2}} \right) \cr
& {f_x}\left( {x,y} \right) = 3{x^2}{e^{3y}} \cr
& \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^3}{e^{3y}}} \right] \cr
& {f_y}\left( {x,y} \right) = {x^3}\frac{\partial }{{\partial y}}\left[ {{e^{3y}}} \right] \cr
& {\text{solving derivatives}} \cr
& {f_y}\left( {x,y} \right) = {x^3}\left( {3{e^{3y}}} \right) \cr
& {f_y}\left( {x,y} \right) = 3{x^3}{e^{3y}} \cr} $$
