Answer
$${f_{xy}}\left( {x,y} \right) = {f_{yx}}\left( {x,y} \right) = - {e^x}\sin y$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {e^x}\cos y \cr
& \cr
& {\text{Find the first derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right) \cr
& \cr
& {\text{Calculate }}{f_x}\left( {x,y} \right){\text{ differentiating with respect to }}x;{\text{ treat }}y{\text{ as a constant}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left( {{e^x}\cos y} \right) \cr
& {f_x}\left( {x,y} \right) = \cos y\frac{\partial }{{\partial x}}\left( {{e^x}} \right) \cr
& {f_x}\left( {x,y} \right) = {e^x}\cos y \cr
& \cr
& {\text{Calculate }}{f_y}\left( {x,y} \right){\text{ differentiating with respect to }}y;{\text{ treat }}x{\text{ as a constant}} \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left( {{e^x}\cos y} \right) \cr
& {f_y}\left( {x,y} \right) = {e^x}\frac{\partial }{{\partial y}}\left( {\cos y} \right) \cr
& {f_y}\left( {x,y} \right) = - {e^x}\sin y \cr
& \cr
& {\text{Find the mixed second - partial derivatives }}{f_{xy}}\left( {x,y} \right){\text{ and }}{f_{yx}}\left( {x,y} \right){\text{ and}} \cr
& {\text{confirm that they are the same}} \cr
& {\text{Calculate }}{f_{xy}}\left( {x,y} \right){\text{ differentiating }}{f_x}\left( {x,y} \right){\text{with respect to }}y. \cr
& {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left( {{e^x}\cos y} \right) \cr
& {f_{xy}}\left( {x,y} \right) = {e^x}\left( { - \sin y} \right) \cr
& {f_{xy}}\left( {x,y} \right) = - {e^x}\sin y \cr
& \cr
& {\text{Calculate }}{f_{yx}}\left( {x,y} \right){\text{ differentiating }}{f_y}\left( {x,y} \right){\text{with respect to }}x. \cr
& {f_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left( { - {e^x}\sin y} \right) \cr
& {f_{yx}}\left( {x,y} \right) = - \sin y\frac{\partial }{{\partial x}}\left( {{e^x}} \right) \cr
& {f_{yx}}\left( {x,y} \right) = - {e^x}\sin y \cr
& \cr
& {\text{Then}}{\text{,}} \cr
& {f_{xy}}\left( {x,y} \right) = {f_{yx}}\left( {x,y} \right) = - {e^x}\sin y \cr} $$