Mathematica's command DSolve cannot solve your ODE in closed form; so, apparently, a closed-form solution does not exist.
However, it will shown here that, in line with your conjecture, for each real $c$ your ODE has solutions $y$ such that $y(x)=(c+o(1))x^2$ as $x\to0+$.
Indeed, your ODE is
\begin{equation*}
x^2 y''+x y'=4(y-x^2y+y^3) \tag{1}
\end{equation*}
for real $x>0$.
Assuming the initial conditions $y(0+)=y'(0+)=0$ and using the Taylor formulas $y'(x)=x\int_0^1 v(sx)\,ds$ and $y(x)=x^2\int_0^1 (1-s)v(sx)\,ds$ with $v:=y''$, we rewrite (1) as
\begin{equation*}
v(x)=\int_0^1 K(x,s)v(sx)\,ds+4x^4\Big(\int_0^1 (1-s)v(sx)\,ds\Big)^3, \tag{2}
\end{equation*}
where
$$K(x,s):=[4(1-x^2)(1-s)-1]s.$$
For any real $c$, setting now
\begin{equation*}
v(x)=c+x w(x),
\end{equation*}
we rewrite (2) as
\begin{equation*}
w(x)=F(w)(x):=-2cx+\int_0^1 K(x,s)w(sx)\,ds+4x^3\Big(\int_0^1 (1-s)(c+xw(sx))\,ds\Big)^3. \tag{3}
\end{equation*}
Take now any real $m>0$ and $h>0$, and let
\begin{equation*}
W_{m,h}:=\{w\in C[0,h]\colon\|w\|\le m\},
\end{equation*}
where $\|w\|:=\max_{x\in[0,h]}|w(x)|$.
Since $|K(x,s)|$ is convex in $x^2$, for $x\in[0,1]$ we have
\begin{equation*}
\int_0^1 |K(x,s)|\,ds\le\max\Big(\int_0^1 |K(0,s)|\,ds,\int_0^1 |K(1,s)|\,ds\Big)
=\max(19/48,1/2)=1/2.
\end{equation*}
So, for any $w\in W_{m,h}$
\begin{equation*}
\|F(w)\|\le2|c|h+m/2+h^3(|c|+hm)^3/2\le m \tag{4}
\end{equation*}
if $h>0$ is small enough -- which will be assumed in what follows. So, $F$ maps $W_{m,h}$ into $W_{m,h}$.
Moreover, if for some $w$ and $u$ in $W_{m,h}$ we have $\|w-u\|\le t$ for some real $t\ge0$, then similarly to (4) we get
\begin{equation*}
\|F(w)-F(u)\|\le t/2+3h^3(|c|+hm)^2ht/2\le\tfrac23\,t
\end{equation*}
if $h>0$ is small enough. So, $F$ is a contractive map of $W_{m,h}$ into $W_{m,h}$, and hence $F$ has a fixed point. That is, equation (3) has a solution $w\in W_{m,h}$.
Thus, ODE (1) has a solution $y$ with $y''(x)=c+x w(x)$ for $x\in[0,h]$ and $w\in W_{m,h}$, and with $y(0+)=y'(0+)=0$. So, $y(x)=cx^2/2+O(x^3)$ as $x\to0+$.
$\Box$
