Solution
= Solution
For a nonzero solution, the <separable differential equation> gives $dy/y=3x^2dx$. Integrating yields $\ln|y|=x^3+C$, hence $y=Ke^{x^3}$. The <initial condition> fixes $K=3/e$, so
$$
\boxed{y(x)=3e^{x^3-1},\qquad x\in\mathbb R.}
$$
The zero solution omitted by division by $y$ cannot satisfy the given <initial condition>. Differentiating the displayed solution verifies $y'=3x^2y$.