Answer
${g_x}\left( {x,y} \right) = 3{x^2}\sin y{\text{ and }}{g_y}\left( {x,y} \right) = {x^3}\cos y$
Work Step by Step
$$\eqalign{
& g\left( {x,y} \right) = {x^3}\sin y \cr
& {\text{Differentiate with respect to }}x \cr
& {g_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{x^3}\sin y} \right] \cr
& {\text{Considering }}y{\text{ as a constant}} \cr
& {g_x}\left( {x,y} \right) = \sin y\frac{\partial }{{\partial x}}\left[ {{x^3}} \right] \cr
& {\text{Differentiating}} \cr
& {g_x}\left( {x,y} \right) = \sin y\left( {3{x^2}} \right) \cr
& {g_x}\left( {x,y} \right) = 3{x^2}\sin y \cr
& \cr
& {\text{Differentiate with respect to }}y \cr
& {g_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^3}\sin y} \right] \cr
& {\text{Considering }}x{\text{ as a constant}} \cr
& {g_y}\left( {x,y} \right) = {x^3}\frac{\partial }{{\partial x}}\left[ {\sin y} \right] \cr
& {\text{Differentiating}} \cr
& {g_y}\left( {x,y} \right) = {x^3}\left( {\cos y} \right) \cr
& {g_y}\left( {x,y} \right) = {x^3}\cos y \cr} $$
