$y' = x \exp (y - x^2)$
$\dfrac{dy}{dx} = x e^{y - x^2}$
$\dfrac{dy}{dx} = x e^y e^{-x^2}$
$\dfrac{dy}{e^y} = x e^{-x^2} \, dx$
$e^{-y} \, dy = e^{-x^2}(x \, dx)$
$\displaystyle -\int e^{-y} \, (-dy) = -\frac{1}{2} \int e^{-x^2}(-2x \, dx)$
$-e^{-y} = -\frac{1}{2} e^{-x^2} - c$
$e^{-y} = \frac{1}{2} e^{-x^2} + c$
when x = 0, y = 0
$e^{-0} = \frac{1}{2} e^{-0^2} + c$
$1 = \frac{1}{2} + c$
$c = \frac{1}{2}$
thus,
$e^{-y} = \frac{1}{2} e^{-x^2} + \frac{1}{2}$
$e^{-y} = \frac{1}{2} e^{-x^2} + \frac{1}{2}$
$\dfrac{1}{e^y} = \dfrac{e^{-x^2} + 1}{2}$
$\dfrac{2}{e^{-x^2} + 1} = e^y$
$\ln \dfrac{2}{e^{-x^2} + 1} = \ln e^y$
$\ln \dfrac{2}{e^{-x^2} + 1} = y \ln e$
$y = \ln \dfrac{2}{e^{-x^2} + 1}$
$y = \ln \dfrac{2}{1 + \exp (-x^2)}$ answer
