Answer
$$\eqalign{
& {h_{xx}}\left( {x,y} \right) = 10y,\,\,\,\,{h_{yy}}\left( {x,y} \right) = 24x \cr
& {h_{xy}}\left( {x,y} \right) = {h_{yx}}\left( {x,y} \right) = 10x + 24y \cr} $$
Work Step by Step
$$\eqalign{
& h\left( {x,y} \right) = 30y + 5{x^2}y + 12x{y^2} \cr
& {\text{Find }}{h_x}\left( {x,y} \right){\text{ and }}{h_y}\left( {x,y} \right) \cr
& {h_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {30y + 5{x^2}y + 12x{y^2}} \right] \cr
& {\text{treat }}y{\text{ as a constant and }}x{\text{ as a variable}}{\text{. then}} \cr
& {h_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {30y} \right] + \frac{\partial }{{\partial x}}\left[ {5{x^2}y} \right] + \frac{\partial }{{\partial x}}\left[ {12x{y^2}} \right] \cr
& {h_x}\left( {x,y} \right) = 10xy + 12{y^2} \cr
& \cr
& {h_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {30y + 5{x^2}y + 12x{y^2}} \right] \cr
& {\text{treat }}x{\text{ as a constant and }}y{\text{ as a variable}}{\text{. then}} \cr
& {h_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {30y} \right] + \frac{\partial }{{\partial y}}\left[ {5{x^2}y} \right] + \frac{\partial }{{\partial y}}\left[ {12x{y^2}} \right] \cr
& {h_y}\left( {x,y} \right) = 30 + 5{x^2} + 24xy \cr
& \cr
& {\text{find the second - order partial derivatives}} \cr
& {h_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {10xy + 12{y^2}} \right] \cr
& {h_{xy}}\left( {x,y} \right) = 24y \cr
& \cr
& {h_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {30 + 5{x^2} + 24xy} \right] \cr
& {h_{yx}}\left( {x,y} \right) = 24y \cr
& \cr
& {h_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {10xy + 12{y^2}} \right] \cr
& {h_{xx}}\left( {x,y} \right) = 10y \cr
& {\text{and}} \cr
& {h_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {30 + 5{x^2} + 24xy} \right] \cr
& {h_{yy}}\left( {x,y} \right) = 24x \cr} $$