Answer
$$\eqalign{
& {f_x}\left( {x,y} \right) = - {y^2}\left( {2 - 6x} \right)\sin \left( {2x{y^2} - 3{x^2}{y^2}} \right) \cr
& {f_y}\left( {x,y} \right) = - \sin \left( {2x{y^2} - 3{x^2}{y^2}} \right)\left( {4xy - 6{x^2}y} \right) \cr} $$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \cos \left( {2x{y^2} - 3{x^2}{y^2}} \right) \cr
& \cr
& {\text{Calculate }}{f_x}\left( {x,y} \right){\text{ differentiating with respect to }}x;{\text{ treat }}y{\text{ as a constant}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\cos \left( {2x{y^2} - 3{x^2}{y^2}} \right)} \right] \cr
& {\text{Use the rule }}\frac{d}{{dx}}\left[ {\cos u} \right] = - \sin u\frac{{du}}{{dx}} \cr
& {f_x}\left( {x,y} \right) = - \sin \left( {2x{y^2} - 3{x^2}{y^2}} \right)\frac{\partial }{{\partial x}}\left[ {2x{y^2} - 3{x^2}{y^2}} \right] \cr
& {f_x}\left( {x,y} \right) = - \sin \left( {2x{y^2} - 3{x^2}{y^2}} \right)\left( {2{y^2} - 6x{y^2}} \right) \cr
& {f_x}\left( {x,y} \right) = - {y^2}\left( {2 - 6x} \right)\sin \left( {2x{y^2} - 3{x^2}{y^2}} \right) \cr
& \cr
& {\text{Calculate }}{f_y}\left( {x,y} \right){\text{ differentiating with respect to }}y;{\text{ treat }}x{\text{ as a constant}} \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\cos \left( {2x{y^2} - 3{x^2}{y^2}} \right)} \right] \cr
& {\text{Use the rule }}\frac{d}{{dy}}\left[ {\cos u} \right] = - \sin u\frac{{du}}{{dy}} \cr
& {f_y}\left( {x,y} \right) = - \sin \left( {2x{y^2} - 3{x^2}{y^2}} \right)\frac{\partial }{{\partial y}}\left[ {2x{y^2} - 3{x^2}{y^2}} \right] \cr
& {f_y}\left( {x,y} \right) = - \sin \left( {2x{y^2} - 3{x^2}{y^2}} \right)\left( {4xy - 6{x^2}y} \right) \cr} $$
