Answer
$$y = \ln \left( {{e^x} + C} \right)$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = \frac{{{e^x}}}{{{e^y}}} \cr
& {\text{Separating variables leads to}} \cr
& {e^y}dy = {e^x}dx \cr
& {\text{To solve this equation}}{\text{, determine the antiderivative of each side}} \cr
& \int {{e^y}dy} = \int {{e^x}dx} \cr
& {\text{integrating by using }}\int {{e^u}du} = {e^u} + C{\text{ and the power rule }} \cr
& {e^y} = {e^x} + C \cr
& {\text{multiply both sides by }} - 1 \cr
& {e^y} = {e^x} + C \cr
& {\text{solve the equation for }}y \cr
& \ln {e^y} = \ln \left( {{e^x} + C} \right) \cr
& y = \ln \left( {{e^x} + C} \right) \cr
& \cr
& {\text{verifying that the solution satisfies the original differential equation}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\ln \left( {{e^x} + C} \right)} \right] \cr
& \frac{{dy}}{{dx}} = \frac{{{e^x}}}{{{e^x} + C}} \cr
& {\text{where }}{e^y} = {e^x} + C.{\text{ then}} \cr
& \frac{{dy}}{{dx}} = \frac{{{e^x}}}{{{e^y}}} \cr
& {\text{The general solution is verified}} \cr} $$
