Differential equation $y' ^2+(y-1)y'-y=0$

The given equation can be written as $(y'+y)(y'-1)=0$, so either $y'=-y$ or $y'=1$, so $y=Ce^{-x}$ and $y=x+C$ are solutions.

Another solution is a piecewise combination of the above. At any point where we switch between the two forms, we have $-y = y' = 1$, so $Ce^{-x}=-1$. This can only occur if $C<0$, at $x=\log(-C)=c$, giving the following additional solutions:

$$y(x) = \begin{cases} -e^{c-x} & x \le c \\ x-c-1 & x \ge c \end{cases}$$ and $$y(x) = \begin{cases} x-c-1 & x \le c \\ -e^{c-x} & x \ge c \end{cases}$$

$$y=\frac{y'(y'-1)}{-(y'-1)}=-y'$$ Thus $y=c e^{-x}$ for some $c\in \mathbb R$.