I am trying to understand a passage in the introduction to this book, which deals with algorithmical procedures to analytically solve ODEs. Specifically, I do not understand how the ODE $$ y''(y+x) + y'(y'-1)=0, \quad y=y(x)$$ (formula (1.1) in the linked page) is transformed into $$ 2v'' u + (v')^2 - 1 =0 \quad v=v(u)$$ by the point transformation $$ x=u+v \qquad y=u-v.$$ This should be easy. However, the fact that the transformation is not fiber-preserving is confusing to me. (By fiber-preserving I mean that the independent variable $x$ is not a function of the independent variable $u$ alone).
Also, the given reference is in German and I cannot read that language.
EDIT. I found a good explanation of the theoretical aspects of this kind of transformations in Olver's book Applications of Lie groups to differential equations, §2.2 "Groups and differential equations".
We have $\displaystyle u=\frac{x+y}{2}$, $\displaystyle v=\frac{x-y}{2}$. Then $\displaystyle \frac{dv}{du}=\frac{\partial v}{\partial x}\frac{dx}{du}+\frac{\partial v}{\partial y}\frac{dy}{du}=\frac{1}{1+y'}-\frac{1}{1+\frac{1}{y'}}=\frac{2}{y'+1}-1$, $\displaystyle \frac{d^2v}{du^2}=\frac{-2}{(y'+1)^2}\frac{dy'}{du}=\frac{-2}{(y'+1)^2}y''\frac{dx}{du}=\frac{-4}{(y'+1)^2}y''\frac{1}{1+y'}=\frac{-4}{(y'+1)^3}y''=\frac{-2}{(y'+1)^3}\frac{y'(1-y')}{u}\implies 2v''u=\frac{-4}{(y'+1)^3}y'(1-y')$.
Now just make the proper substitutions for $y'$ from $\displaystyle \frac{dv}{du}$.