Answer
$${f_{xy}}\left( {x,y} \right) = {\text{ }}{f_{yx}}\left( {x,y} \right) = - \sin x$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = \cos xy \cr
& {\text{Find the first partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right){\text{ then}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\cos xy} \right] \cr
& {\text{treat }}y{\text{ as a constant}}{\text{, then}} \cr
& {f_x}\left( {x,y} \right) = y\frac{\partial }{{\partial x}}\left[ {\cos x} \right] \cr
& {f_x}\left( {x,y} \right) = - y\sin x \cr
& and \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {\cos xy} \right] \cr
& {\text{treat }}x{\text{ as a constant}}{\text{, then }} \cr
& {f_y}\left( {x,y} \right) = \cos x\frac{\partial }{{\partial y}}\left[ y \right] \cr
& {f_y}\left( {x,y} \right) = \cos x \cr
& \cr
& {\text{Find the second partial derivatives }}{f_{xy}}\left( {x,y} \right){\text{ and }}{f_{yx}}\left( {x,y} \right){\text{ then}} \cr
& {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ { - y\sin x} \right] \cr
& {f_{xy}}\left( {x,y} \right) = - \sin x\frac{\partial }{{\partial y}}\left[ y \right] \cr
& {f_{xy}}\left( {x,y} \right) = - \sin x \cr
& and \cr
& {\text{ }}{f_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {\cos x} \right] \cr
& {\text{ }}{f_{yx}}\left( {x,y} \right) = - \sin x \cr
& \cr
& {\text{Then}}{\text{, we can verify that }}{f_{xy}}\left( {x,y} \right) = {\text{ }}{f_{yx}}\left( {x,y} \right) = - \sin x \cr} $$
