<s>[INST] Solve differential equation $yy'(yy'-2x)=x^2-2y^2$. 
$$yy'(yy'-2x)=x^2-2y^2$$  

I've tried to divide by $y^2$ and substitude $\frac{x}{y}=z$, but it led to:
$$y'^2-2zy'=z^2-2$$
$$(y'-z)^2=2(z^2-1)$$
$$|y'-z|=\sqrt{2}\sqrt{z^2-1}$$ hence
$y'=\sqrt{2}\sqrt{z^2-1}+z$ or $y'=-\sqrt{2}\sqrt{z^2-1}+z$ and I don't know any way to integrate it. Is there some hint for the very beginning maybe that I don't know?
 [/INST] With $y^2=u$ then $2yy'=u'$ so
\begin{align}
2yy'(2yy'-4x)&=4x^2-8y^2 \\
u'(u'-4x)+4x^2&=8x^2-8u \\
(u'-2x)^2&=8(x^2-u) \\
-d(x^2-u)&=2\sqrt{2}\sqrt{x^2-u^2}dx \\
\sqrt{x^2-u}&=C-\sqrt{2}x \\
x^2-y^2=&(C-\sqrt{2}x)^2
\end{align}
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