Solve $2\ddot{y}y - 3(\dot{y})^2 + 8x^2 = 0$

Let $z=e^{ky}$. Then $z'=ky'e^{ky}$ and $$z''=ky''e^{ky}+k^2(y')^2 e^{ky}\text.$$

If we put $k=-\frac{3}{2}$ then $z''=-\frac{3}{2}y''e^{ky}+\frac{3}{2}^2(y')^2 e^{ky}$, or $\frac{2}{3}z''=-y''e^{ky}+\frac{3}{2}(y')^2 e^{ky}$, or $$-\frac{4}{3}z''=2y''e^{ky}-3(y')^2 e^{ky}\text{, or even}$$ $$-\frac{4z''}{3z}=2y''-3(y')^2\text.$$

Your original equation is now $$\frac{4z''}{3z}=8x^2\text,$$ which I hope will get you closer.

Indeed...

Given $z''=6x^2z$, expanding $z$ as a power series in $x$ yields $a_n=\frac{a_{n-2}}{(n+1)(n+2)}$ times a constant, which makes one think that $z$ must be something to do with $e^{x^2}$.

Put $$z=e^{ax^2+bx+c}$$ and differentiate twice, and you get something like $$z''=2az+4a^2x^2z+2abxz+b^2z\text,$$ which is enticing but has an embarrassing $xz$ term. Fortunately that is the only term in which $b$ isn't squared. So if you take $$z=e^{ax^2+bx+c}+e^{ax^2-bx+c}$$ and differentiate it twice, you will get rid of the unwanted $xz$ term and you will have constraints on $a$ and $b$ which make $z$ satisfy the differential equation.

General note: All this has been done on the back of one and a half envelopes, so do check it.

$$ \begin{cases} 3y''y + 3(y')^2 - 2x^2 = 0, \\ y(0) = 1, \\ y'(0) = 0. \end{cases} $$

$$y''y+y'^2=(y'y)'\quad\to\quad 3(y'y)'=2x^2$$ $$3y'y=\frac{2}{3}x^3+c_1$$ $y'(0)=0\quad\to\quad c_1=0$ $$y'y=\frac{2}{9}x^3$$ $$2y'y=(y^2)'=\frac{4}{9}x^3$$ $$y^2=\frac{1}{9}x^4+c_2$$ $y(0)=1\quad\to\quad c_2=1$ $$y=\sqrt{\frac{1}{9}x^4+1}$$