= Solution
Construct the <scalar potential> $F(x,y)=ye^{x^2}-x^2$. Its differential is
$$
dF=2x(ye^{x^2}-1)dx+e^{x^2}dy,
$$
which proves exactness directly. Multiplying the <differential equation> by $e^{x^2}$ shows that a solution satisfies $dF/dx=0$, so $ye^{x^2}-x^2=C$. The initial value at zero gives $C=-1$. Therefore
$$
\boxed{y(x)=(x^2-1)e^{-x^2},\qquad x\in\mathbb R.}
$$
The factor $e^{x^2}$ is also an <integrating factor> for the associated first-order linear equation.
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