Answer
$ |\ln y | -\dfrac{y^2}{2}=\dfrac{x^2}{2} +C$
Work Step by Step
Write the differential equation as:
$\dfrac{dy}{dx}=\dfrac{xy}{(1-y^2)x}$
or, $ \dfrac{1-y^2}{y} \ dy = x \ dx$
or, $ \int x \ dx =\int \dfrac{1-y^2}{y} \ dy $
or, $\int \dfrac{1}{y} \ dy -\int y \ dy =\int x dx$
or, $|\ln y | -\dfrac{y^2}{2}=\dfrac{x^2}{2} +C$
