The ODE $$xy'^2-yy'-x=0$$ See: Solve $xy'^{~2}-yy'-x=0$ for a new family of intersecting self-orthogonal trajectories
and the ODE $$xy'^2-2yy'-x=0$$
See: prove $x^2 = 4 c(y+c)$ is self orthogonal trajectory
are invariant under $y'\rightarrow -\frac{1}{y'}$ and give rise to two families of intersecting self-orthogonal trajectories.
Similarly, the simpler ODE $$y'^2-2xy'-1=0$$ can also give rise to such a family of trajectories but with an interesting difference with regard to the two examples referred here. The question is to find this family of curve(s) and their disparate feature.
As noted by @MOO
$$y'^2-2xy'-1=0 \implies y'= x\pm \sqrt{1+x^2} \implies \int dy= \int(x \pm \sqrt{x^2+1}) dx$$ $$ \implies y= \frac{x^2}{2}\pm \left(\frac{x\sqrt{1+x^2}}{2}+\sinh^{-1} x \right)+C~~~(1)$$
So for a fixed value of $C$, these are two curves that cut each other orthogonally. Also for all real values of $C$, these curves represent a network of self-orthogonal curves.
This situation is different from that of earlier two examples mention in the OP where a single curve represents self-orthogonal trajectories for $C>0$ and $C<0$, for instance, the confocal parabolas: $y^2=4C(x+C).$ See the orhohonal trajecories for $C=1,2,3,4$ (blue) and $C=-1,-2,-3,-4$ (red) in the fig. below
below.
Similarly, teachers may also motivate students by the family of intersecting self-orthogonal trajectories arising from $y'^2=1$, these are $y=\pm x+C$. It may be mentioned that $\frac{x^2}{a^2+C}+\frac{y^2}{b^2+C}=1$ for fixed values of $a$ and $b$ is a family of nonintersecting-self-orthogonal curves.