Answer
$${g_{xx}}\left( {x,y} \right) = 60{x^2}{y^2},\,\,\,\,{g_{yy}}\left( {x,y} \right) = 40{x^4} + 72y,\,\,\,\,{g_{xy}}\left( {x,y} \right) = \,{g_{yx}}\left( {x,y} \right) = 140{x^3}y$$
Work Step by Step
$$\eqalign{
& g\left( {x,y} \right) = 5{x^4}{y^2} + 12{y^3} - 9x \cr
& {\text{Find }}{g_x}\left( {x,y} \right){\text{ and }}{g_y}\left( {x,y} \right) \cr
& {g_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {5{x^4}{y^2} + 12{y^3} - 9x} \right] \cr
& {\text{treat }}y{\text{ as a constant and }}x{\text{ as a variable}}{\text{. then}} \cr
& {g_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {5{x^4}{y^2}} \right] + \frac{\partial }{{\partial x}}\left[ {12{y^3}} \right] - \frac{\partial }{{\partial x}}\left[ {9x} \right] \cr
& {g_x}\left( {x,y} \right) = 20{x^3}{y^2} - 9 \cr
& \cr
& {g_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {5{x^4}{y^2} + 12{y^3} - 9x} \right] \cr
& {\text{treat }}x{\text{ as a constant and }}y{\text{ as a variable}}{\text{. then}} \cr
& {g_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {5{x^4}{y^2}} \right] + \frac{\partial }{{\partial y}}\left[ {12{y^3}} \right] - \frac{\partial }{{\partial y}}\left[ {9x} \right] \cr
& {g_y}\left( {x,y} \right) = 10{x^4}y + 36{y^2} \cr
& \cr
& {\text{find the second - order partial derivatives}} \cr
& {g_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {20{x^3}{y^2} - 9} \right] \cr
& {g_{xy}}\left( {x,y} \right) = 40{x^3}y \cr
& \cr
& {g_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {10{x^4}y + 36{y^2}} \right] \cr
& {g_{yx}}\left( {x,y} \right) = 40{x^3}y \cr
& \cr
& {g_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {20{x^3}{y^2} - 9} \right] \cr
& {g_{xx}}\left( {x,y} \right) = 60{x^2}{y^2} \cr
& {\text{and}} \cr
& {g_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {10{x^4}y + 36{y^2}} \right] \cr
& {g_{yy}}\left( {x,y} \right) = 40{x^4} + 72y \cr} $$