Answer
$$\eqalign{
& {h_{xy}}\left( {x,y} \right) = \cos y - \sin x \cr
& {h_{yx}}\left( {x,y} \right) = \cos y - \sin x \cr} $$
Work Step by Step
$$\eqalign{
& h\left( {x,y} \right) = x\sin y + y\cos x \cr
& {\text{Calculate }}{h_x}\left( {x,y} \right){\text{ treating }}y{\text{ as a constant}} \cr
& {h_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {x\sin y + y\cos x} \right] \cr
& {h_x}\left( {x,y} \right) = \sin y - y\sin x \cr
& {\text{Calculate }}{h_{xy}}\left( {x,y} \right){\text{ differentiating }}{h_x}\left( {x,y} \right){\text{ with respect to }}y \cr
& {\text{and treating }}y{\text{ as a constant}}{\text{.}} \cr
& {h_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\sin y - y\sin x} \right] \cr
& {h_{xy}}\left( {x,y} \right) = \cos y - \sin x \cr
& \cr
& {\text{Calculate }}{h_y}\left( {x,y} \right){\text{ treating }}x{\text{ as a constant}} \cr
& {h_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {x\sin y + y\cos x} \right] \cr
& {h_y}\left( {x,y} \right) = x\cos y + \cos x \cr
& {\text{Calculate }}{h_{yx}}\left( {x,y} \right){\text{ differentiating }}{h_y}\left( {x,y} \right){\text{ with respect to }}x \cr
& {\text{and treating }}y{\text{ as a constant}}{\text{.}} \cr
& {h_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {x\cos y + \cos x} \right] \cr
& {h_{yx}}\left( {x,y} \right) = \cos y - \sin x \cr} $$
