Answer
$$\eqalign{
& {f_x}\left( {x,y} \right) = 15{x^2} \cr
& {f_y}\left( {x,y} \right) = 7 \cr} $$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 5{x^3} + 7y - 3 \cr
& {\text{Calculate }}{f_x}\left( {x,y} \right){\text{ treating }}y{\text{ as a constant}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {5{x^3} + 7y - 3} \right] \cr
& {f_x}\left( {x,y} \right) = 5\left( {3{x^2}} \right) + 0 - 0 \cr
& {f_x}\left( {x,y} \right) = 15{x^2} \cr
& {\text{Calculate }}{f_y}\left( {x,y} \right){\text{ treating }}x{\text{ as a constant}} \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {5{x^3} + 7y - 3} \right] \cr
& {f_y}\left( {x,y} \right) = 0 + 7\left( 1 \right) - 0 \cr
& {f_y}\left( {x,y} \right) = 7 \cr} $$
