Differential equation: $y=2xy'+\frac{1}{(y')^2}$ I've encountered the following differential equation:
$y=2xy'+\frac{1}{(y')^2}$
I tried to differentiate in order to $x$ and then used $p=y'$. After some calculation, I arrived at the formula $ x = \frac{2}{p^3} + \frac{C_{1}}{p^2}$
Does anyone has any suggestion to solve the first equation?
Thanks for your attention.
 A: This is a Lagrange Equation given as:  $~y=x\phi(p)+\psi(p).$ where $p=\frac{dy}{dx}$
Given equation can be written as

$$y=2xp+1/p^2      ~~~~~~~~~~~.......(1)$$

On diff. wrt $x$,
$\Rightarrow dy=2xdp+2pdx-\frac{2}{p^3}dp$ 
$\Rightarrow -pdx=(2x-\frac{2}{p^3})dp~~~~~$  [I have used $dy=p dx$]
$\Rightarrow \frac{dx}{dp}+\frac{2x}{p}=\frac{2}{p^4};~~~~~~~~~~~~~~~$   [Assuming $p \ne 0$]
Above Eq. is linear in $x$ and $p$ variable.
Hence its solution in parametric form is given as 

$$~x(p)= \frac{-2}{p^3}+\frac{C}{p^2}$$

From Eq(1),

$$y(p)= \frac{-3}{p^2}+\frac{2C}{p}$$ 

Singular Solution SS is given by $p=0$, Putting this back into Eq.(1),
We get the SS as

$$y(x)=0$$

A: By eliminating parameter $p$, I find this (cartesian) equation for the singular solution:

$$y=3x^{2/3}$$

Besides, let us look for a particular power solution of the differential equation :
$$y=ax^b$$
(which is a different issue). There exist exactly one such solution.
Indeed, we must have, for any $x \neq 0$ :
$$ax^b=2xabx^{b-1}+\dfrac{1}{(abx^{b-1})^2}\tag{1}$$
Multiplying LHS and RHS by $x^{-b}$, (1) is equivalent to :
$$a(1-2b)=\dfrac{1}{a^2b^2}x^{2-3b}\tag{2}$$
(2) cannot hold for any $x$ unless $2-3b=0$, and simultaneously $a^3b^2(1-2b)=1$. 
This system of equations has a unique solution : 
$$b=\dfrac23  \ \ \text{and} \ \ a=-\dfrac{3}{\sqrt[3]{4}}$$
Therefore, the differential equation has this power solution 

$$y=-\dfrac{3}{\sqrt[3]{4}}x^{2/3}\tag{3}$$

It corresponds to the case $C=0$ of the parametric equations given in the general solutions by @Axion004 and @math. 
