Answer
$${h_x}\left( { - 2,1} \right) = - 4{\text{ and }}{h_y}\left( { - 2,1} \right) = - 2$$
Work Step by Step
$$\eqalign{
& h\left( {x,y} \right) = {x^2} - {y^2},{\text{ }}\left( { - 2,1,3} \right) \cr
& {\text{Calculate the partial derivatives }}{h_x}\left( {x,y} \right){\text{ and }}{h_y}\left( {x,y} \right) \cr
& {h_x}\left( {x,y} \right) = \frac{d}{{dx}}\left[ {{x^2} - {y^2}} \right] \cr
& {h_x}\left( {x,y} \right) = 2x \cr
& {\text{Evaluate at }}\left( { - 2,1,3} \right) \cr
& {h_x}\left( { - 2,1} \right) = 2\left( { - 2} \right) \cr
& {h_x}\left( { - 2,1} \right) = - 4 \cr
& and \cr
& {h_y}\left( {x,y} \right) = \frac{d}{{dy}}\left[ {{x^2} - {y^2}} \right] \cr
& {h_y}\left( {x,y} \right) = - 2y \cr
& {\text{Evaluate at }}\left( { - 2,1,3} \right) \cr
& {h_y}\left( { - 2,1} \right) = - 2\left( 1 \right) \cr
& {h_y}\left( { - 2,1} \right) = - 2 \cr} $$