The solution of the differential equation $\frac{dy}{dx}+\frac{3x^2}{1+x^3}y=\frac{{\sin }^{2}x}{1+x^3}$ is:

$\frac{dy}{dx}+\frac{3x^2}{1+x^3}y=\frac{{\sin }^{2}x}{1+x^3}$

Here, $P=\frac{3x^2}{1+x^3}\Rightarrow \text{ I.F. }=e^{\int Pdx}=e^{\log \left(1+x^3\right)}=1+x^3$

Thus the solution is $y\cdot \left(1+x^3\right)=\int \frac{{\sin }^{2}x}{1+x^3}\left(1+x^3\right)dx$

$\begin{array}{c}=\int \frac{1-\cos 2x}{2}dx\\ \Rightarrow y\left(1+x^3\right)=\frac{1}{2}x-\frac{\sin 2x}{4}+c\end{array}$