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