Answer
$${h_{xx}}\left( {x,y} \right) = 6x,\,\,\,\,\,\,\,\,{\text{ }}{h_{yy}}\left( {x,y} \right) = 2x,\,\,\,\,\,\,\,{h_{xy}}\left( {x,y} \right) = 2y{\text{ and }}{h_{yx}}\left( {x,y} \right) = 2y$$
Work Step by Step
$$\eqalign{
& h\left( {x,y} \right) = {x^3} + x{y^2} + 1 \cr
& {\text{Find the first partial derivatives }}{h_x}\left( {x,y} \right){\text{ and }}{h_y}\left( {x,y} \right){\text{ then}} \cr
& {h_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{x^3} + x{y^2} + 1} \right] \cr
& {\text{treat }}y{\text{ as a constant}}{\text{, then}} \cr
& {h_x}\left( {x,y} \right) = 3{x^2} + {y^2} \cr
& and \cr
& {h_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{x^3} + x{y^2} + 1} \right] \cr
& {\text{treat }}x{\text{ as a constant}}{\text{, then }} \cr
& {h_y}\left( {x,y} \right) = 2xy \cr
& \cr
& {\text{Find the second partial derivatives }}{h_{xy}}\left( {x,y} \right){\text{ and }}{h_{yx}}\left( {x,y} \right){\text{ then}} \cr
& {h_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {3{x^2} + {y^2}} \right] \cr
& {h_{xy}}\left( {x,y} \right) = 2y \cr
& and \cr
& {\text{ }}{h_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {2xy} \right] \cr
& {\text{ }}{h_{yx}}\left( {x,y} \right) = 2y \cr
& \cr
& {\text{Find the second partial derivatives }}{h_{xx}}\left( {x,y} \right){\text{ and }}{h_{yy}}\left( {x,y} \right){\text{ then}} \cr
& {h_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {3{x^2} + {y^2}} \right] \cr
& {h_{xx}}\left( {x,y} \right) = 6x \cr
& and \cr
& {\text{ }}{h_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {2xy} \right] \cr
& {\text{ }}{h_{yy}}\left( {x,y} \right) = 2x \cr} $$
