Answer
$f(x)=\frac{e^{2x}(1+2x)}{1-e^{-x}}$
Work Step by Step
Differentiate both sides and use the $FTC$ it follows:
$$f(x)=(xe^{2x})'+e^{-x}f(x)$$
$$f(x)=e^{2x}+2xe^{2x}+e^{-x}f(x)$$
$$f(x)-e^{-x}f(x)=e^{2x}+2xe^{2x}$$
$$f(x)-e^{-x}f(x)=e^{2x}(1+2x)$$
$$f(x)(1-e^{-x})=e^{2x}(1+2x)$$
$$f(x)=\frac{e^{2x}(1+2x)}{1-e^{-x}}$$
