Answer
$$\eqalign{
& {f_x}\left( {x,y} \right) = {e^x}\cos y \cr
& {f_y}\left( {x,y} \right) = - {e^x}\sin y \cr} $$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {e^x}\cos y \cr
& {\text{Calculate }}{f_x}\left( {x,y} \right){\text{ treating }}y{\text{ as a constant}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{e^x}\cos y} \right] \cr
& {f_x}\left( {x,y} \right) = \cos y\frac{\partial }{{\partial x}}\left[ {{e^x}} \right] \cr
& {f_x}\left( {x,y} \right) = {e^x}\cos y \cr
& {\text{Calculate }}{f_y}\left( {x,y} \right){\text{ treating }}x{\text{ as a constant}} \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{e^x}\cos y} \right] \cr
& {f_y}\left( {x,y} \right) = {e^x}\frac{\partial }{{\partial y}}\left[ {\cos y} \right] \cr
& {f_y}\left( {x,y} \right) = {e^x}\left( { - \sin y} \right) \cr
& {f_y}\left( {x,y} \right) = - {e^x}\sin y \cr} $$
