Answer
$$y=e^{-x}+1+Ce^{e^{-x}}$$
Work Step by Step
Rewrite the equation as
$$y'+e^{-x}y=e^{-2x}$$
This is a linear equation which has the integrating factor as follows
$$\alpha(x)= e^{\int P(x)dx}=e^{\int e^{-x}dx}=e^{-e^{-x}} .$$
Now the general solution is
\begin{align} y&=\alpha^{-1}(x)\left( \int\alpha(x) Q(x)dx +C\right)\\ &=e^{e^{-x}}\left( \int e^{-2x} e^{-e^{-x}} dx +C\right)\\ &=e^{e^{-x}} \left(e^{-x} e^{-e^{-x}} +e^{-e^{-x}} +C\right) .\end{align}
Where we did integration by parts.
Thus, the general solution is
$$y=e^{e^{-x}} \left(e^{-x} e^{-e^{-x}} +e^{-e^{-x}} +C\right) .$$
We can simplify this as:
$$y=e^{-x}+1+Ce^{e^{-x}}$$
