Answer
$$\eqalign{
& {f_{xx}}\left( {x,y} \right) = 6y,\,\,\,{f_{xy}}\left( {x,y} \right) = 6x + 3{y^2} \cr
& {f_{yy}}\left( {x,y} \right) = 6xy,\,\,\,{f_{yx}}\left( {x,y} \right) = 6x + 3{y^2} \cr} $$
Work Step by Step
$$\eqalign{
& {\text{Let }}f\left( {x,y} \right) = 3{x^2}y + x{y^3} \cr
& {\text{Calculate }}{f_{xx}}\left( {x,y} \right){\text{ and }}{f_{xy}}\left( {x,y} \right) \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {3{x^2}y + x{y^3}} \right] \cr
& {f_x}\left( {x,y} \right) = 6xy + {y^3} \cr
& {f_{xx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {6xy + {y^3}} \right] \cr
& {f_{xx}}\left( {x,y} \right) = 6y \cr
& and \cr
& {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {6xy + {y^3}} \right] \cr
& {f_{xy}}\left( {x,y} \right) = 6x + 3{y^2} \cr
& \cr
& {\text{Calculate }}{f_{yy}}\left( {x,y} \right){\text{ and }}{f_{yx}}\left( {x,y} \right) \cr
& {\text{using the product rule}} \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {3{x^2}y + x{y^3}} \right] \cr
& {f_y}\left( {x,y} \right) = 3{x^2} + 3x{y^2} \cr
& {f_{yy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {3{x^2} + 3x{y^2}} \right] \cr
& {f_{yy}}\left( {x,y} \right) = 6xy \cr
& and \cr
& {f_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {3{x^2} + 3x{y^2}} \right] \cr
& {f_{yx}}\left( {x,y} \right) = 6x + 3{y^2} \cr} $$
