Solve the differential equation $6y^2 y'^2= y - 3y'x$ I want to solve the DFE $6y^2 y'^2= y - 3y'x$, but am stuck at the end. This is what I did:
$$6y^2 y'^2= y - 3y'x \Leftrightarrow x = \frac{6y^2 y'^2 - y}{-3y'} = \frac{y}{3y'} - 2y^2y'.$$ Suppose now $dy/ dx = y' = p$ then $p' = dp/dy$ (if we take the derivative with respect to $y$) and $dx/dy = 1/p$. 
So now we have $$x = \frac{y}{3p} - 2y^2 p.$$ Taking the derivative with respect to $y$ yields $$\frac{1}{p} = \frac{1}{3p} - \frac{1}{3p^2}p'y - 4yp - 2y^2p'=  \frac{1}{3p} - 4yp - \left(\frac{1}{3p^2}y + 2y^2\right)p'.$$
But now I don't know how to go on. Any tips?
 A: Your equation can be solved by the method on p. 36 of Hildebrand, "Advanced Calculus for Applications" (1976), ISBN 0-13-011189-9. Here is that method applied to your equation:
First, notice that the equation is quadratic in $y^\prime$. Solve to find
$$\frac{dy}{dx}=\frac{-3x\pm\sqrt{9x^2+24y^3}}{12y^2}\,.$$
This suggests the substitution $u = 9x^2+24y^3$. Calculate the derivative of $u$ with respect to $x$: $$\frac{du}{dx} = 18x + 72y^2\frac{dy}{dx}\,.$$ Using the earlier expression for $\frac{dy}{dx}$, we therefore have
$$\frac{du}{dx} = 18x + 6\left(-3x \pm \sqrt{9x^2+24y^3}\right)$$
$$\frac{du}{dx} = \pm 6\sqrt{u}\,.$$
These equations can be solved by separation of variables:
$$u(x) = 9x^2 \pm 6cx + c^2\,.$$
Comparing to $u(x) = 9x^2 + 24y^3$, we have the solution in the form
$$24y^3 = c^2\pm 6cx$$
or $$y(x) = \frac{1}{24^{1/3}}(c^2\pm 6cx)^{1/3}$$ where $c$ is any constant.
A: Alternatively, one can find that multiplying the original equation with $3y^2$ allows to substitute $u=y^3$ to get
$$
2u'^2=3u-3u'x\iff u=u'x+\frac23u'^2 .
$$
This is a Clairaut equation with the lines $$u=cx+\frac23c^2$$ as solution family and $$u'=-\frac34x\implies u=-\frac38x^2$$ as envelope.
