Answer
$${f_{xy}}\left( {x,y} \right) = {\text{ }}{f_{yx}}\left( {x,y} \right) = 0$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 2{x^3} + 3{y^2} + 1 \cr
& {\text{Find the first partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right){\text{ then}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {2{x^3} + 3{y^2} + 1} \right] \cr
& {\text{treat }}y{\text{ as a constant}}{\text{, then}} \cr
& {f_x}\left( {x,y} \right) = 6{x^2} \cr
& and \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {2{x^3} + 3{y^2} + 1} \right] \cr
& {\text{treat }}x{\text{ as a constant}}{\text{, then }} \cr
& {f_y}\left( {x,y} \right) = 6{y^2} \cr
& \cr
& {\text{Find the second partial derivatives }}{f_{xy}}\left( {x,y} \right){\text{ and }}{f_{yx}}\left( {x,y} \right){\text{ then}} \cr
& {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {6{x^2}} \right] \cr
& {f_{xy}}\left( {x,y} \right) = 0 \cr
& and \cr
& {\text{ }}{f_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {6{y^2}} \right] \cr
& {\text{ }}{f_{yx}}\left( {x,y} \right) = 0 \cr
& \cr
& {\text{Then}}{\text{, we can verify that }}{f_{xy}}\left( {x,y} \right) = {\text{ }}{f_{yx}}\left( {x,y} \right) = 0 \cr} $$