Answer
${g_x}\left( {x,y} \right) = 3\left( {2x + y} \right){\left( {{x^2} + xy} \right)^2}{\text{ and }}{g_y}\left( {x,y} \right) = 3x{\left( {{x^2} + xy} \right)^2}$
Work Step by Step
$$\eqalign{
& g\left( {x,y} \right) = {\left( {{x^2} + xy} \right)^3} \cr
& {\text{Differentiate with respect to }}x \cr
& {g_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{{\left( {{x^2} + xy} \right)}^3}} \right] \cr
& {\text{Using the Chain Rule}} \cr
& {g_x}\left( {x,y} \right) = 3{\left( {{x^2} + xy} \right)^2}\frac{\partial }{{\partial x}}\left[ {{x^2} + xy} \right] \cr
& {\text{Considering }}y{\text{ as a constant}} \cr
& {g_x}\left( {x,y} \right) = 3{\left( {{x^2} + xy} \right)^2}\left( {2x + y} \right) \cr
& {g_x}\left( {x,y} \right) = 3\left( {2x + y} \right){\left( {{x^2} + xy} \right)^2} \cr
& \cr
& {\text{Differentiate with respect to }}y \cr
& {g_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{{\left( {{x^2} + xy} \right)}^3}} \right] \cr
& {\text{Using the Chain Rule}} \cr
& {g_y}\left( {x,y} \right) = 3{\left( {{x^2} + xy} \right)^2}\frac{\partial }{{\partial y}}\left[ {{x^2} + xy} \right] \cr
& {\text{Considering }}x{\text{ as a constant}} \cr
& {g_y}\left( {x,y} \right) = 3{\left( {{x^2} + xy} \right)^2}\left( {0 + x} \right) \cr
& {g_y}\left( {x,y} \right) = 3x{\left( {{x^2} + xy} \right)^2} \cr} $$
