Derivative of square of derivative? I was trying to solve this differential equation:
$$2yy'' + 3y'^2 = 4y^2 $$
And I found this way to solver it: http://eqworld.ipmnet.ru/en/solutions/ode/ode0344.pdf but I don't understand why $w'_y = y''_{xx}$. If $w(y) = (y'_x)^2$, how can I find this:
$$ \dfrac{d}{dy}\bigg(\dfrac{dy}{dx}\bigg)^2$$
 A: By the Chain Rule, 
$$\frac{d}{dy}\left(\frac{dy}{dx}\right)^2=\frac{dx}{dy}\frac{d}{dx}\left(\frac{dy}{dx}\right)^2.$$
Now use the fact that $\dfrac{dx}{dy}=\dfrac{1}{\dfrac{dy}{dx}}.$
Calculate $\dfrac{d}{dx}\left(\dfrac{dy}{dx}\right)^2$ using the Product Rule. When we put things together, there is some nice cancellation, which undoubtedly means there is a simple conceptual reason. 
A: Here's something that can't possibly be right:
$$\frac{dw}{dy} = \frac{dw}{dx}\frac{dx}{dy} = \frac{2y'y''}{\frac{dy}{dx}} = 2y'' $$
The funny thing is this will achieve the result given by the reference if we add $y''+f(y)(y')^2 + g(y) = 0$ to itself and make the substitution.
A: I'll give this a shot.  Let $z=\frac{dy}{dx}$
$$\frac{dw}{dx}=\frac{dz^2}{dy}=2z\frac{dz}{dy}=2z\times\frac{\frac{dz}{dx}}{\frac{dy}{dx}}=2z\times\frac{(\frac{d^2y}{dx^2})}z=2\frac{d^2y}{dx^2}$$
This appears to reduce your equation to
$$yw'+3w=4y^2$$
$$w'+\frac{3w}y=4y$$
A: Let's set $\ w(y):=(y_x')^2\ $ then :
$$\frac {dw(y)}{dx}=\frac {dw(y)}{dy}\frac {dy}{dx}=\frac {d\left(\left(\frac {dy}{dx}\right)^2\right)}{dx}=2\frac {dy}{dx}\frac {d^2y}{dx^2}$$
From the second and fourth term we get (if $\frac {dy}{dx}\not = 0$) :
$$\frac {dw(y)}{dy}=2\frac {d^2y}{dx^2}$$
