Solve $\frac{dy}{dx}=e^{x-y}(e^x-e^y)$

Multiply by $e^y$:
$$e^y\frac{dy}{dx}=e^x(e^x-e^y)⟹e^y\frac{dy}{dx}+e^x{e}^y=e^{2x}$$
Let $v=e^y⟹\frac{dv}{dx}=e^y\frac{dy}{dx}$ $
$\frac{dv}{dx}+e^{x}v=e^{2x}$$IntegratingFactor$ IF = e^{\int e^x dx} = e^{e^x}$Solution:$ v \cdot e^{e^x} = \int e^{2x} \cdot e^{e^x} dx$Let$ t = e^x \implies dt = e^x dx
$\int t e^t dt = t e^t - e^t + c
$$e^y\cdot e^{e^x}=e^x{e}^{e^x}-e^{e^x}+c$$
Divide by $e^{e^x}$:
$$e^y=e^x-1+c{e}^{-e^x}$$