Answer
$${g_x}\left( {1,1} \right) = - 2{\text{ and }}{g_y}\left( {1,1} \right) = - 2$$
Work Step by Step
$$\eqalign{
& g\left( {x,y} \right) = 4 - {x^2} - {y^2},{\text{ }}\left( {1,1,2} \right) \cr
& {\text{Calculate the partial derivatives }}{g_x}\left( {x,y} \right){\text{ and }}{g_y}\left( {x,y} \right) \cr
& {g_x}\left( {x,y} \right) = \frac{d}{{dx}}\left[ {4 - {x^2} - {y^2}} \right] \cr
& {g_x}\left( {x,y} \right) = - 2x \cr
& {\text{Evaluate at }}\left( {1,1,2} \right) \cr
& {g_x}\left( {1,1} \right) = - 2\left( 1 \right) \cr
& {g_x}\left( {1,1} \right) = - 2 \cr
& and \cr
& {g_y}\left( {x,y} \right) = \frac{d}{{dy}}\left[ {4 - {x^2} - {y^2}} \right] \cr
& {g_y}\left( {x,y} \right) = - 2y \cr
& {\text{Evaluate at }}\left( {1,1,2} \right) \cr
& {g_y}\left( {1,1} \right) = - 2\left( 1 \right) \cr
& {g_y}\left( {1,1} \right) = - 2 \cr} $$
