Let, $2x+y=t\Rightarrow \frac{dy}{dx}+2=\frac{dt}{dx}$
$\frac{dt}{dx}+xt=x^3{t}^3\Rightarrow \frac{1}{t^3}\frac{dt}{dx}+\frac{1}{t^2}x=x^3$
Let, $\frac{1}{t^2}=u\Rightarrow \frac{-2}{t^3}\frac{dt}{dx}=\frac{du}{dx}$
$\frac{du}{dx}+\left(-2x\right)u=-2x^3$
$\mathrm{I.F.}=e^{-\int 2xdx}=e^{-x^2}\Rightarrow u.e^{-x^2}=\int e^{-x^2}\left(-2x^3\right)dx$
$\frac{e^{-x^2}}{{\left(2x+y\right)}^2}=-2\int e^{-x^2}.x^{3}dx$
$\frac{e^{-x^2}}{{\left(2x+y\right)}^2}=\int e^{-x^2}.x^2\left(-2x\right)dx$

Let, $-x^2=v$
$$\begin{gathered} -2xdx=dv \\ \Rightarrow \frac{e^{-x^2}}{{(2x+y)}^2}=-\int e^{v}vdv \end{gathered}$$
$$\begin{gathered} \frac{e^{-x^2}}{{(2x+y)}^2}+v\cdot e^v-e^v=C \\ \Rightarrow \frac{e^{-x^2}}{{(2x+y)}^2}-x^2{e}^{-x^2}-e^{-x^2}=C \end{gathered}$$
$\frac{1}{{(2x+y)}^2}=\left(x^2+1\right)+C{\mathrm{e}}^{x^2}$