I was trying to solve this exercise, $yy''=3(y')^2$ The question asks to reduce the ODE to first-order before proceeding to solve the equation. For your information, I am still in Homogeneous Linear ODE and I find it hard to find the solution. So I decided to check the solution on Slader (Quizlet + now). However, it creates more confusion. This is the first part of the solution in Slader.
$z'=\frac{d^{2}y}{dx^2}=\frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}(z(y))=\frac{dz}{dy}\frac{dy}{dx}=z'\cdot y'$
Assuming $z=y’$.
I couldn't wrap my head around this part. $\frac{d}{dx}(z(y))$ It is very confusing.
You could directly separate to get logarithmic derivatives $$ \frac{y''}{y'}=3\frac{y'}{y}. $$
The trick that is proposed to you works better backwards, assume that $y$ is in bijection with $x$ so that the free parameter can be changed, so that $y'(x)=z(y(x))$. Now apply the chain rule for the second derivative, $y''(x)=z'(y(x))y'(x)=z'(y)z(y)$. If you insert that you get $$ z'(y)z(y)y=3z(y)^2\implies z(y)·(z'(y)y-3z(y))=0. $$ As the segments under consideration have $z(y)=y'\ne 0$, one can extract the second factor and solve