Answer
$$y=\frac{C_1e^{x^{2} / 2}}{\sqrt{1+x^{2}}} $$
Work Step by Step
Given
$$\left(1+x^{2}\right) y'=x^{3} y $$
Then
\begin{align*}
\left(1+x^{2}\right) \frac{d y}{d x}&=x^{3} y\\
\frac{1}{y} d y&=\frac{x^{3}}{1+x^{2}} d x
\end{align*}
Hence
\begin{aligned} \int \frac{1}{y} d y &=\int\left(x-\frac{x}{1+x^{2}}\right) d x \\ \ln y &=\frac{x^{2}}{2}-\frac{1}{2} \ln \left(1+x^{2}\right)+C \\ y &=\frac{C_1e^{x^{2} / 2}}{\sqrt{1+x^{2}}} \end{aligned}
