= Solution
Away from $x=-1$ the <separable differential equation> gives
$$
\frac{dy}{y}=\frac{x^2}{1+x^3}dx,\qquad
\ln|y|=\frac13\ln|1+x^3|+C.
$$
On the <interval> containing $x=1$, one has $x>-1$, so the positive initial value selects $y=K(1+x^3)^{1/3}$. Substitution of $y(1)=2$ gives $K=2^{2/3}$. Thus
$$
\boxed{y(x)=2^{2/3}(1+x^3)^{1/3},\qquad x>-1.}
$$
The zero solution omitted by division by $y$ cannot satisfy the initial value. Although the selected solution tends to zero as $x\downarrow-1$, its <derivative> diverges there; it does not extend through that point as a classical <differentiable> solution.
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