Answer
$$y = \ln \left( {x + 1} \right)$$
Work Step by Step
$$\eqalign{
& {\text{Let the differential equation }}\frac{{dy}}{{dx}} = {e^{ - y}} \cr
& {\text{Separate the variables}} \cr
& {e^y}dy = dx \cr
& {\text{Integrate both sides with respect to }}x \cr
& {e^y} = x + C\,\,\left( {\bf{1}} \right) \cr
& \cr
& {\text{Find the function }}y = f\left( x \right){\text{ with }}y = 0{\text{ when }}x = 0 \cr
& {e^0} = 0 + C \cr
& C = 1 \cr
& \cr
& {\text{Substitute }}C = 1{\text{ into equation }}\left( {\bf{1}} \right) \cr
& {e^y} = x + 1\,\, \cr
& {\text{Solving for }}y \cr
& y = \ln \left( {x + 1} \right) \cr} $$