= Solution
Since $1+x^2>0$ for all real $x$, divide the equation by that factor. Its linear coefficient can be integrated by <polynomial division>:
$$
P(x)=\frac{3x^3}{1+x^2}=3x-\frac{3x}{1+x^2},\qquad
\int P(x)\,dx=\frac32x^2-\frac32\ln(1+x^2).
$$
Thus a positive <integrating factor> is $M=e^{3x^2/2}(1+x^2)^{-3/2}$. Multiplying the divided equation gives the <product rule> form
$$
\frac d{dx}(My)=\frac{6x}{(1+x^2)^{5/2}}.
$$
The <integration by substitution> $u=1+x^2$ gives
$$
My=-\frac{2}{(1+x^2)^{3/2}}+C,\qquad
y=e^{-3x^2/2}\big[C(1+x^2)^{3/2}-2\big].
$$
The <initial condition> $y(0)=1$ fixes $C-2=1$. The required solution is therefore
$$
\boxed{y(x)=e^{-3x^2/2}\left[3(1+x^2)^{3/2}-2\right],\qquad x\in\mathbb R.}
$$
No real singularity was introduced by the division or the <integrating factor>, so the solution extends over the entire real line.
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