Answer
$${f_{xy}}\left( {x,y} \right) = {\text{ }}{f_{yx}}\left( {x,y} \right) = - 6x{y^{ - 2}} + 4{x^{ - 2}}y$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 3{x^2}{y^{ - 1}} - 2{x^{ - 1}}{y^2} \cr
& {\text{Find the first partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right){\text{ then}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {3{x^2}{y^{ - 1}} - 2{x^{ - 1}}{y^2}} \right] \cr
& {\text{treat }}y{\text{ as a constant}}{\text{, then}} \cr
& {f_x}\left( {x,y} \right) = {y^{ - 1}}\frac{\partial }{{\partial x}}\left[ {3{x^2}} \right] - 2{y^2}\frac{\partial }{{\partial x}}\left[ {{x^{ - 1}}} \right] \cr
& {f_x}\left( {x,y} \right) = 6x{y^{ - 1}} + 2{x^{ - 2}}{y^2} \cr
& and \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {3{x^2}{y^{ - 1}} - 2{x^{ - 1}}{y^2}} \right] \cr
& {\text{treat }}x{\text{ as a constant}}{\text{, then }} \cr
& {f_y}\left( {x,y} \right) = 3{x^2}\frac{\partial }{{\partial y}}\left[ {{y^{ - 1}}} \right] - 2{x^{ - 1}}\frac{\partial }{{\partial y}}\left[ {{y^2}} \right] \cr
& {f_y}\left( {x,y} \right) = - 3{x^2}{y^{ - 2}} - 4{x^{ - 1}}y \cr
& \cr
& {\text{Find the second partial derivatives }}{f_{xy}}\left( {x,y} \right){\text{ and }}{f_{yx}}\left( {x,y} \right){\text{ then}} \cr
& {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {6x{y^{ - 1}} + 2{x^{ - 2}}{y^2}} \right] \cr
& {f_{xy}}\left( {x,y} \right) = - 6x{y^{ - 2}} + 4{x^{ - 2}}y \cr
& and \cr
& {\text{ }}{f_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ { - 3{x^2}{y^{ - 2}} - 4{x^{ - 1}}y} \right] \cr
& {\text{ }}{f_{yx}}\left( {x,y} \right) = - 6x{y^{ - 2}} + 4{x^{ - 2}}y \cr
& \cr
& {\text{Then}}{\text{, we can verify that }}{f_{xy}}\left( {x,y} \right) = {\text{ }}{f_{yx}}\left( {x,y} \right) = - 6x{y^{ - 2}} + 4{x^{ - 2}}y \cr} $$
