Answer
$${f_{xy}}\left( {x,y} \right) = {f_{yx}}\left( {x,y} \right) = - 32{y^3}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 4{x^2} - 8x{y^4} + 7{y^5} - 3 \cr
& {\text{Find the first derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& \cr
& {\text{Calculate }}{f_x}\left( {x,y} \right){\text{ differentiating with respect to }}x;{\text{ treat }}y{\text{ as a constant}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left( {4{x^2} - 8x{y^4} + 7{y^5} - 3} \right) \cr
& {f_x}\left( {x,y} \right) = 8x - 8{y^4} \cr
& \cr
& {\text{Calculate }}{f_y}\left( {x,y} \right){\text{ differentiating with respect to }}y;{\text{ treat }}x{\text{ as a constant}} \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left( {4{x^2} - 8x{y^4} + 7{y^5} - 3} \right) \cr
& {f_y}\left( {x,y} \right) = - 32x{y^3} \cr
& \cr
& {\text{Find the mixed second - partial derivatives }}{f_{xy}}\left( {x,y} \right){\text{ and }}{f_{yx}}\left( {x,y} \right){\text{ and}} \cr
& {\text{confirm that they are the same}} \cr
& {\text{Calculate }}{f_{xy}}\left( {x,y} \right){\text{ differentiating }}{f_x}\left( {x,y} \right){\text{with respect to }}y. \cr
& {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left( {8x - 8{y^4}} \right) \cr
& {f_{xy}}\left( {x,y} \right) = - 32{y^3} \cr
& \cr
& {\text{Calculate }}{f_{yx}}\left( {x,y} \right){\text{ differentiating }}{f_y}\left( {x,y} \right){\text{with respect to }}x. \cr
& {f_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left( { - 32x{y^3}} \right) \cr
& {f_{yx}}\left( {x,y} \right) = - 32{y^3} \cr
& \cr
& {\text{Then}}{\text{,}} \cr
& {f_{xy}}\left( {x,y} \right) = {f_{yx}}\left( {x,y} \right) = - 32{y^3} \cr} $$
