Answer
$$y = \ln \left( {{e^x} + 2} \right)$$
Work Step by Step
$$\eqalign{
& \frac{{dy}}{{dx}} = {e^{x - y}} \cr
& \frac{{dy}}{{dx}} = \frac{{{e^x}}}{{{e^y}}} \cr
& {\text{separate}} \cr
& {e^y}dy = {e^x}dx \cr
& {\text{integrate both sides}} \cr
& \int {{e^y}} dy = \int {{e^x}} dx \cr
& {e^y} = {e^x} + C \cr
& {\text{the initial condition }}y\left( 0 \right) = \ln 3{\text{ implies that}} \cr
& {e^{\ln 3}} = {e^0} + C \cr
& 3 = 1 + C \cr
& C = 2 \cr
& {e^y} = {e^x} + 2 \cr
& {\text{solve for }}y \cr
& y = \ln \left( {{e^x} + 2} \right) \cr} $$
