Answer
$y=-\dfrac{2}{\ln (x^2+1)+2}$
Work Step by Step
We are given that $ \dfrac{dy}{dx}=\dfrac{xy^2}{x^2+1}$
We will separate the variables to obtain:
$\dfrac{dy}{y^2}=\dfrac{x \ dx}{x^2+1}$
Integrate to obtain:
$\int \dfrac{dy}{y^2}=\int \dfrac{x \ dx}{x^2+1} \ dx$
This implies that $\dfrac{-1}{y}=\dfrac{\ln (x^2+1)}{2}+C$
After applying the initial conditions, $y=-1$ when $x=0$, we get $C=1$
Therefore, we have: $\dfrac{-1}{y}=\dfrac{\ln (x^2+1)}{2}+1 \implies \dfrac{-1}{y}=\dfrac{\ln (x^2+1)+2}{2}$
or, $y=-\dfrac{2}{\ln (x^2+1)+2}$
