Answer
$g_x(x,y) = 5y(2xy + 1)(x + {x^2}y)^4$
$g_y(x,y) = (x + x^2y)^4 (6x^2y+x)$
Work Step by Step
$$\eqalign{
& g\left( {x,y} \right) = y{\left( {x + {x^2}y} \right)^5} \cr
& \cr
& {\text{Differentiate with respect to }}x \cr
& {g_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {y{{\left( {x + {x^2}y} \right)}^5}} \right] \cr
& {\text{Considering }}y{\text{ as a constant}} \cr
& {g_x}\left( {x,y} \right) = y\frac{\partial }{{\partial x}}\left[ {{{\left( {x + {x^2}y} \right)}^5}} \right] \cr
& {\text{Using the Chain Rule}} \cr
& {g_x}\left( {x,y} \right) = 5y{\left( {x + {x^2}y} \right)^4}\frac{\partial }{{\partial x}}\left[ {x + {x^2}y} \right] \cr
& {g_x}\left( {x,y} \right) = 5y{\left( {x + {x^2}y} \right)^4}\left( {1 + 2xy} \right) \cr
& {g_x}\left( {x,y} \right) = 5y\left( {2xy + 1} \right){\left( {x + {x^2}y} \right)^4} \cr
& \cr
& {\text{Differentiate with respect to }}y \cr
& {g_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {y{{\left( {x + {x^2}y} \right)}^5}} \right] \cr
& {\text{Use the product rule together the chain rule}} \cr
& {g_y}\left( {x,y} \right) = y\frac{\partial }{{\partial y}}\left[ {{{\left( {x + {x^2}y} \right)}^5}} \right] + {\left( {x + {x^2}y} \right)^5}\frac{\partial }{{\partial y}}\left[ y \right] \cr
& {g_y}\left( {x,y} \right) = 5y{\left( {x + {x^2}y} \right)^4}\left( {0 + {x^2}} \right) + {\left( {x + {x^2}y} \right)^5}\left( 1 \right) \cr
& {\text{Simplify}} \cr
& {g_y}\left( {x,y} \right) = 5{x^2}y{\left( {x + {x^2}y} \right)^4} + {\left( {x + {x^2}y} \right)^5} \cr
&{g_y}\left( {x,y} \right) =(x+x^2y)^4(5x^2y+x+x^2y)\cr
&{g_y}\left( {x,y} \right) =(x+x^2y)^4(6x^2y+x)} $$