How do I solve this differential equation $y''^2-2y'y''+3=0$ in parametric form? I do have quite no idea about this one.
The obvious substitution $y'=p, y''=p\frac{dp}{dy}$ doesn't make the situation any better

Let $z=y'$, then $$\frac{dz}{dx}=z\pm\sqrt{z^2-3}\\ \frac{dx}{dz}=\frac{z\pm\sqrt{z^2-3}}3$$
then solve $x$ as a function of $z$
Sadly, the final integral, to get $y$ as a function of $x$, looks difficult.
