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?

$$yy'(yy'-2x)=x^2-2y^2$$
divide by $y^2$:
 $$(y')^2-2\frac{x}{y}y'+(2-(\frac{x}{y})^2)=0$$
$$y'=\frac{x}{y}±\sqrt {2(\frac{x}{y})^2-2}$$
substitute $y=vx$
$$v+x\frac{dv}{dx}=\frac{1}{v}±\sqrt {2(\frac{1}{v})^2-2}$$
$$\frac{dx}{x}=\frac{dv}{\frac{1}{v}-v±\sqrt {2(\frac{1}{v})^2-2}}$$
$$\frac{dx}{x}=\frac{vdv}{1-v^2±\sqrt {2-2v^2}}$$
As, $(1-v^2)>0, v=\sin t$ can be substituted
$$\frac{dx}{x}=\frac{\sin t \cos t \space dt}{\cos^2 t+\sqrt {2} \cos t}$$
$$\frac{dx}{x}=\frac{\sin t \space dt}{\cos t+\sqrt {2}}$$
Integrating on both sides: 
$$ln|x|=-ln|\cos t+\sqrt {2}|+c$$
$$x(\cos t+\sqrt {2})=c$$
$$\cos^2 t=(\frac{c}{x}-\sqrt(2))^2$$
$$1-\frac{y}{x}^2=(\frac{c}{x}-\sqrt{2})^2$$
This can be rewritten as:
$$x^2-y^2=(\sqrt{2}\space x-c)^2$$
