Answer
$${f_x}\left( {x,y} \right) = 20{x^3}{y^3} - 30{x^4}y{\text{ and }}{f_y}\left( {x,y} \right) = 15{x^4}{y^2} - 6{x^5}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 5{x^4}{y^3} - 6{x^5}y \cr
& {\text{find }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {5{x^4}{y^3} - 6{x^5}y} \right] \cr
& {\text{treat y as a constant and }}x{\text{ as a variable}}{\text{. then}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {5{x^4}{y^3}} \right] - \frac{\partial }{{\partial x}}\left[ {6{x^5}y} \right] \cr
& {f_x}\left( {x,y} \right) = 5{y^3}\frac{\partial }{{\partial x}}\left[ {{x^4}} \right] - 6y\frac{\partial }{{\partial x}}\left[ {{x^5}} \right] \cr
& {f_x}\left( {x,y} \right) = 5{y^3}\left( {4{x^3}} \right) - 6y\left( {5{x^4}} \right) \cr
& {f_x}\left( {x,y} \right) = 20{x^3}{y^3} - 30{x^4}y \cr
& \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {5{x^4}{y^3} - 6{x^5}y} \right] \cr
& {\text{treat x as a constant and y as a variable}}{\text{. then}} \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {5{x^4}{y^3}} \right] - \frac{\partial }{{\partial y}}\left[ {6{x^5}y} \right] \cr
& {f_y}\left( {x,y} \right) = 5{x^4}\frac{\partial }{{\partial y}}\left[ {{y^3}} \right] - 6{x^5}\frac{\partial }{{\partial y}}\left[ y \right] \cr
& {f_y}\left( {x,y} \right) = 5{x^4}\left( {3{y^2}} \right) - 6{x^5}\left( 1 \right) \cr
& {f_y}\left( {x,y} \right) = 15{x^4}{y^2} - 6{x^5} \cr} $$
