Answer
$$\eqalign{
& {f_x}\left( {x,y} \right) = 8x - 2y \cr
& {f_y}\left( {x,y} \right) = - 2x + 2y \cr} $$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 4{x^2} - 2xy + {y^2} \cr
& {\text{Calculate }}{f_x}\left( {x,y} \right){\text{ treating }}y{\text{ as a constant}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {4{x^2} - 2xy + {y^2}} \right] \cr
& {f_x}\left( {x,y} \right) = 4\left( {2x} \right) - 2y\left( 1 \right) + 0 \cr
& {f_x}\left( {x,y} \right) = 8x - 2y \cr
& {\text{Calculate }}{f_y}\left( {x,y} \right){\text{ treating }}x{\text{ as a constant}} \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {4{x^2} - 2xy + {y^2}} \right] \cr
& {f_y}\left( {x,y} \right) = 0 - 2x\left( 1 \right) + 2y \cr
& {f_y}\left( {x,y} \right) = - 2x + 2y \cr} $$
