Answer
$${f_{xy}}\left( {x,y} \right) = {\text{ }}{f_{yx}}\left( {x,y} \right) = {e^{x + y}}$$
Work Step by Step
$$\eqalign{
& f\left( {x,y} \right) = {e^{x + y}} \cr
& {\text{Find the first partial derivatives }}{f_x}\left( {x,y} \right){\text{ and }}{f_y}\left( {x,y} \right){\text{ then}} \cr
& {f_x}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{e^{x + y}}} \right] \cr
& {\text{treat }}y{\text{ as a constant}}{\text{, then}} \cr
& {f_x}\left( {x,y} \right) = {e^{x + y}}\frac{\partial }{{\partial x}}\left[ {x + y} \right] \cr
& {f_x}\left( {x,y} \right) = {e^{x + y}}\left( 1 \right) \cr
& {f_x}\left( {x,y} \right) = {e^{x + y}} \cr
& and \cr
& {f_y}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{e^{x + y}}} \right] \cr
& {\text{treat }}x{\text{ as a constant}}{\text{, then }} \cr
& {f_y}\left( {x,y} \right) = {e^{x + y}}\frac{\partial }{{\partial y}}\left[ {x + y} \right] \cr
& {f_y}\left( {x,y} \right) = {e^{x + y}}\left( 1 \right) \cr
& {f_y}\left( {x,y} \right) = {e^{x + y}} \cr
& \cr
& {\text{Find the second partial derivatives }}{f_{xy}}\left( {x,y} \right){\text{ and }}{f_{yx}}\left( {x,y} \right){\text{ then}} \cr
& {f_{xy}}\left( {x,y} \right) = \frac{\partial }{{\partial y}}\left[ {{e^{x + y}}} \right] \cr
& {f_{xy}}\left( {x,y} \right) = {e^{x + y}} \cr
& and \cr
& {\text{ }}{f_{yx}}\left( {x,y} \right) = \frac{\partial }{{\partial x}}\left[ {{e^{x + y}}} \right] \cr
& {\text{ }}{f_{yx}}\left( {x,y} \right) = {e^{x + y}} \cr
& \cr
& {\text{Then}}{\text{, we can verify that }}{f_{xy}}\left( {x,y} \right) = {\text{ }}{f_{yx}}\left( {x,y} \right) = {e^{x + y}} \cr} $$