Answer
$${f_x}\left( {x,y} \right) = 3{x^2}{e^{ - y}} + \frac{{{y^3}\sec \sqrt x \tan \sqrt x }}{{2\sqrt x }},\,\,\,\,\,\,{f_y}\left( {x,y} \right) = - {x^3}{e^{ - y}} + 3{y^2}\sec \sqrt x $$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {x^3}{e^{ - y}} + {y^3}\sec \sqrt x \cr
& {\text{Calculate }}{f_x}\left( {x,y} \right){\text{ differentiating with respect to }}x;{\text{ treat }}y{\text{ as a constant}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left( {{x^3}{e^{ - y}} + {y^3}\sec \sqrt x } \right) \cr
& {\text{Solving derivatives}}{\text{,}}\,\,\,{\text{use power rule and }}\frac{d}{{dx}}\left[ {\sec u} \right] = \sec u\tan u\frac{{du}}{{dx}} \cr
& {f_x}\left( {x,y} \right) = 3{x^2}{e^{ - y}} + {y^3}\sec \sqrt x \tan \sqrt x \left( {\frac{1}{{2\sqrt x }}} \right) \cr
& {f_x}\left( {x,y} \right) = 3{x^2}{e^{ - y}} + \frac{{{y^3}\sec \sqrt x \tan \sqrt x }}{{2\sqrt x }} \cr
& \cr
& {\text{Calculate }}{f_y}\left( {x,y} \right){\text{ differentiating with respect to }}y;{\text{ treat }}x{\text{ as a constant}} \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left( {{x^3}{e^{ - y}} + {y^3}\sec \sqrt x } \right) \cr
& {\text{Solving derivatives}} \cr
& {f_y}\left( {x,y} \right) = {x^3}\left( { - {e^{ - y}}} \right) + \sec \sqrt x \left( {3{y^2}} \right) \cr
& {f_y}\left( {x,y} \right) = - {x^3}{e^{ - y}} + 3{y^2}\sec \sqrt x \cr} $$
