Answer
$${f_x}\left( {x,y} \right) = 18x{y^2},\,\,\,\,{f_y}\left( {x,y} \right) = 18{x^2}y - 8y,\,\,\,\,\,{f_x}\left( {2, - 1} \right) = 36{\text{ and }}{f_y}\left( { - 4,3} \right) = 840$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 9{x^2}{y^2} - 4{y^2} \cr
& {\text{Find }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {9{x^2}{y^2} - 4{y^2}} \right] \cr
& {\text{treat y as a constant and }}x{\text{ as a variable}}{\text{. then}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {9{x^2}{y^2}} \right] - \frac{\partial }{{\partial x}}\left[ {4{y^2}} \right] \cr
& {f_x}\left( {x,y} \right) = 9{y^2}\left( {2x} \right) \cr
& {f_x}\left( {x,y} \right) = 18x{y^2} \cr
& {\text{evaluate }}{f_x}\left( {2, - 1} \right) \cr
& {f_x}\left( {2, - 1} \right) = 18\left( 2 \right){\left( { - 1} \right)^2} \cr
& {f_x}\left( {2, - 1} \right) = 36 \cr
& \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {9{x^2}{y^2} - 4{y^2}} \right] \cr
& {\text{treat x as a constant and y as a variable}}{\text{. then}} \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {9{x^2}{y^2}} \right] - \frac{\partial }{{\partial y}}\left[ {4{y^2}} \right] \cr
& {f_y}\left( {x,y} \right) = 9{x^2}\left( {2y} \right) - 8y \cr
& {f_y}\left( {x,y} \right) = 18{x^2}y - 8y \cr
& {\text{evaluate }}{f_y}\left( { - 4,3} \right) \cr
& {f_y}\left( { - 4,3} \right) = 18{\left( { - 4} \right)^2}\left( 3 \right) - 8\left( 3 \right) \cr
& {f_y}\left( { - 4,3} \right) = 864 - 24 \cr
& {f_y}\left( { - 4,3} \right) = 840 \cr} $$
