Answer
$y=\dfrac{e^x}{e^x-xe^x+c}$
Work Step by Step
On compare the equation $y'-y=xy^2$ with Bernoulli with $n=2$
Suppose $p=y^{1-2}=y^{-1}$
Then $\dfrac{dp}{dx}=(-1)y^{-2}\dfrac{dy}{dx}$
$\dfrac{dp}{dx}+(\dfrac{1}{y}) =-x \implies \dfrac{dp}{dx}+p=x$
The integrating factor is: $I=e^{\int (1)dx}=e^{x}$
$\dfrac{dp}{dx}+p=x \implies \int [p(e^x)]'=\int xe^x dx $
Thus, $-pe^x=e^{x}(x-1)+c \implies y=\dfrac{e^x}{e^x-(xe^x)+c}$
