$y' = y - xy^3e^{-2x}$
$\dfrac{dy}{dx} - y = -xe^{-2x}y^3$
$dy - y~dx = -xe^{-2x}y^3~dx$ → Bernoulli's equation
$dy + Py~dx = Qy^n~dx$
From which
$P = -1$
$Q = -xe^{-2x}$
$n = 3$
$(1 - n) = -2$
$z = y^{1 - n} = y^{-2}$
Integrating factor,
$u = e^{(1 - n)\int P~dx} = e^{-2\int (-1)~dx}$
$u = e^{2\int dx} = e^{2x}$
Thus,
$\displaystyle zu = (1 - n)\int Qu~dx + C$
$\displaystyle y^{-2}(e^{2x}) = -2\int (-xe^{-2x})(e^{2x})~dx + C$
$\displaystyle e^{2x}y^{-2} = 2\int x~dx + C$
$\dfrac{e^{2x}}{y^2} = x^2 + c$
$e^{2x} = y^2(x^2 + c)$ answer