Answer
$${f_x}\left( {3,1} \right) = - 6,\,\,\,{\text{and }}\,\,\,{f_y}\left( {3,1} \right) = - 21$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = 9 - {x^2} - 7{y^3};\,\,\,\,\,\,\,{f_x}\left( {3,1} \right),\,\,\,\,{f_y}\left( {3,1} \right) \cr
& \cr
& {\text{Calculate }}{f_x}\left( {x,y} \right){\text{ differentiating with respect to }}x;{\text{ treat }}y{\text{ as a constant}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left( {9 - {x^2} - 7{y^3}\,\,} \right) \cr
& {f_x}\left( {x,y} \right) = - 2x \cr
& {\text{Evaluating }}{f_x}\left( {3,1} \right) \cr
& {f_x}\left( {3,1} \right) = - 2\left( 3 \right) \cr
& {f_x}\left( {3,1} \right) = - 6 \cr
& \cr
& {\text{Calculate }}{f_y}\left( {x,y} \right){\text{ differentiating with respect to }}y;{\text{ treat }}x{\text{ as a constant}} \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left( {9 - {x^2} - 7{y^3}\,\,} \right) \cr
& {f_y}\left( {x,y} \right) = - 21{y^2} \cr
& {\text{Evaluating }}{f_y}\left( {3,1} \right) \cr
& {f_y}\left( {3,1} \right) = - 21{\left( 1 \right)^2} \cr
& {f_y}\left( {3,1} \right) = - 21 \cr} $$
