Answer
$${f_x}\left( {x,y} \right) = 12x{y^3}{\text{ and }}{f_y}\left( {x,y} \right) = 18{x^2}{y^2} - 4$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 6{x^2}{y^3} - 4y \cr
& {\text{find }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {6{x^2}{y^3} - 4y} \right] \cr
& {\text{treat y as a constant and }}x{\text{ as a variable}}{\text{. then}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {6{x^2}{y^3}} \right] - \frac{\partial }{{\partial x}}\left[ {4y} \right] \cr
& {f_x}\left( {x,y} \right) = 6{y^3}\frac{\partial }{{\partial x}}\left[ {{x^2}} \right] - 4y\frac{\partial }{{\partial x}}\left[ 1 \right] \cr
& {f_x}\left( {x,y} \right) = 12x{y^3} \cr
& \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {6{x^2}{y^3} - 4y} \right] \cr
& {\text{treat x as a constant and y as a variable}}{\text{. then}} \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {6{x^2}{y^3}} \right] - \frac{\partial }{{\partial y}}\left[ {4y} \right] \cr
& {f_y}\left( {x,y} \right) = 6{x^2}\frac{\partial }{{\partial x}}\left[ {{y^3}} \right] - \frac{\partial }{{\partial y}}\left[ {4y} \right] \cr
& {f_y}\left( {x,y} \right) = 6{x^2}\left( {3{y^2}} \right) - 4 \cr
& {f_y}\left( {x,y} \right) = 18{x^2}{y^2} - 4 \cr} $$
