Answer
$$\eqalign{
& {f_{xy}}\left( {x,y} \right) = - 1 \cr
& {f_{yx}}\left( {x,y} \right) = - 1 \cr} $$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 3{x^2} - xy + 2{y^3} \cr
& {\text{Calculate }}{f_x}\left( {x,y} \right){\text{ treating }}y{\text{ as a constant}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {3{x^2} - xy + 2{y^3}} \right] \cr
& {f_x}\left( {x,y} \right) = 6x - y\left( 1 \right) + 0 \cr
& {f_x}\left( {x,y} \right) = 6x - y \cr
& {\text{Calculate }}{f_{xy}}\left( {x,y} \right){\text{ differentiating }}{f_x}\left( {x,y} \right){\text{ with respect to }}y \cr
& {\text{treating }}y{\text{ as a constant}}{\text{.}} \cr
& {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {6x - y} \right] \cr
& {f_{xy}}\left( {x,y} \right) = - 1 \cr
& \cr
& {\text{Calculate }}{f_y}\left( {x,y} \right){\text{ treating }}x{\text{ as a constant}} \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {3{x^2} - xy + 2{y^3}} \right] \cr
& {f_y}\left( {x,y} \right) = - x + 6{y^2} \cr
& {\text{Calculate }}{f_{yx}}\left( {x,y} \right){\text{ differentiating }}{f_y}\left( {x,y} \right){\text{ with respect to }}x \cr
& {\text{and treating }}y{\text{ as a constant}}{\text{.}} \cr
& {f_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ { - x + 6{y^2}} \right] \cr
& {f_{yx}}\left( {x,y} \right) = - 1 \cr} $$
