how to solve $(y')^4 x -2 y (y')^3 + 12 x^3 = 0$

Set $y(x)=\frac12u(x^2)$, then $y'(x)=xu'(x^2)$ and $$ x^5[u'(x^2)]^4-x^3u(x^2)[u'(x^2)]^3+12x^3=0. $$ Cancelling $x^3$, dividing by $y'(x)^3$, substituting $x^2=s$ and reordering gives the form $$ u(s)=su'(s)+12u'(s)^{-3}. $$ This now is a Clairaut DE with linear solution family $$ u(s)=Cs+12C^{-3},~~y(x)=\frac{C}2x^2 +6C^{-3} $$ You get one of the Maple solutions with $C=\frac1{C_1}$.

The singular solution or envelope of $u(s)=su'(s)+f(u'(s))$ is obtained from $0=s+f'(u'(s))$, here $$ 0=s-36[u'(s)]^{-4} $$ The real solutions are $$ u'(s)=\pm\left(\frac{36}{s}\right)^{1/4} $$ so that $$ u(s)=u'(s)\left(s+12[u'(s)]^{-4}\right)=\pm\frac43s\left(\frac{36}{s}\right)^{1/4} \\~\\ y(x)=\pm\frac23x\sqrt{6|x|} $$

This should work for $x>0$ as well as for $x<0$.

As usual, one can switch branches where solutions from the regular family touch the singular solution.

Let's look for polynomial solutions. If $y = c x^d + \ldots$, then $(y')^4 x - 2 y (y')^3 = c^4 (d^4-2 d^3) x^{4d-3} + \ldots$. This $x^{4d-3}$ term can only cancel if $d=2$. So now try $y = c_2 x^2 + c_1 x + c_0$ and it's not hard to see we must have $c_1 = 0$ and $c_0 c_2^3 = 3/4$.

Here's a continuation of what you were doing. The ODE is generalized homogeneous (which you've called isobaric, same thing), so under the transformation $y=x^{3/2}u$, $\ln|x|=\xi$ the ODE becomes \begin{align} \left(u'+\frac{3u}{2}\right)^4-2u\left(u'+\frac{3u}{2}\right)^3+12=0. \end{align} Solving this for $u'$ doesn't work out well, but one can use the method of 'integration by differentiation' outlined in Handbook of exact solutions for ordinary differential equations / Andrei D. Polyanin Valentin F. Zaitsev 2ed. Taking $u'=t$ the equation can be written as \begin{align}\tag{*}\label{1} F(\xi,u,t)=\left(t+\frac{3u}{2}\right)^4-2u\left(t+\frac{3u}{2}\right)^3+12=0. \end{align} Taking the differential of \ref{1} and using the relationship $\mathrm du/\mathrm d\xi=t$ one arrives at the two equations \begin{align} (F_{\xi}+tF_u)\xi'_t+F_t=0,\quad\quad (F_{\xi}+tF_u)u'_t+tF_t=0. \end{align} In our case this yields that \begin{align} \frac{t}{4}(2t+3u)^2(4t-3u)\xi'_t+t(2t+3u)^2&=0 \quad\text{and}\\ \frac{t}{4}(2t+3u)^2(4t-3u)u'_t+t^2(2t+3u)^2&=0. \end{align} (The two equations are identical, via the chain rule we see that $\xi'_t=u'_t/t$). For the second we get that \begin{align} t(2t+3u)^2\big[(4t-3u)u'_t+4t\big]=0. \end{align} The first solution, $t=0$, yields that $y=cx^{3/2}$. Substituting this into the original equation we find that this solves the equation for $c^2=\pm 8/3$. The four different $c$'s correspond to the four particular solutions Maple has given you.

For $2t+3u=0$ we find that $y=c$, which does not solve the equation, I'm not sure why this occurs. For \begin{align} (4t-3u)u'_t+4t=0 \end{align} we can undo some of the substitutions (using $u'_t=u'_\xi/u''_{\xi\xi})$ to get a constant coefficient linear equation \begin{align} u''_{\xi\xi}+u'_{\xi}-\frac{3}{4}u=0. \end{align} This has the solution \begin{align} u(\xi)=c_1e^{\xi/2}+c_2e^{-3\xi/2}\quad\rightarrow\quad y(x)=c_1x^2+c_2. \end{align} Substitution into the original equation yields our general solution \begin{align} y(x)=c_1x^2+\frac{3}{4c_1^3}. \end{align}

Plot the general solution along with the two real particular solutions and you'll see that by changing $c_1$ the general solution arcs along the particular one! Very nice.

My feeling is that this could be extended into the complex domain, but my skills stop at solving ODE's. Perhaps a skilled individual could do so.