Answer
$${e^{ - 2y}} + 2{e^x} = C$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{{{e^{2x + y}}}}{{{e^{x - y}}}} \cr
& {\text{Use the property }}\frac{{{a^m}}}{{{a^n}}} = {a^{m - n}} \cr
& \frac{{dy}}{{dx}} = {e^{2x + y - x + y}} \cr
& \frac{{dy}}{{dx}} = {e^{x + 2y}} \cr
& {\text{Use the property }}{a^{m + n}} = {a^m}{a^n} \cr
& \frac{{dy}}{{dx}} = {e^x}{e^{2y}} \cr
& {\text{Separate the variables}} \cr
& {e^{ - 2y}}dy = {e^x}dx \cr
& {\text{Integrate both sides}} \cr
& \int {{e^{ - 2y}}} dy = \int {{e^x}} dx \cr
& - \frac{1}{2}{e^{ - 2y}} = {e^x} + {C_1} \cr
& {e^{ - 2y}} = - 2{e^x} - 2{C_1} \cr
& {e^{ - 2y}} = - 2{e^x} + C \cr
& {e^{ - 2y}} + 2{e^x} = C \cr} $$
