Solve $xy'^{~2}-yy'-x=0$ for a new family of intersecting self-orthogonal trajectories Self-orthogonal families of curves are rare. One is the family of non-intersecting  concentric ellipses: $$\frac{x^2}{a^2+\lambda}+\frac{y^2}{b^2+\lambda}=1$$ 
The other one is the family of intersecting confocal parabolas:
prove $x^2 = 4 c(y+c)$ is self orthogonal trajectory
The first order ODE: $$xy'^{~2}-y y'-x=0$$  is invariant under $y' \rightarrow -\frac{1}{y'},$ hence its solution would give rise to
a family of self-orthogonal curves. 
The question is: What is the explicit/implicit/parametric form of this new family of curves?
 A: Let us denote $y'=\frac{dy}{dx}$ by $p$, Then the ODE is $$xp^2-yp-x=0 \implies x=\frac{yp}{p^2-1}~~~~(1)$$
Differentiate w.r.t. $y$, to get 
$$\frac{dx}{dy}p^2+2xp\frac{dp}{dy}-p-y\frac{dp}{dy}-\frac{dx}{dy}~~~~(2)$$
$$(2xp-y)\frac{dp}{dy}-\frac{1}{p}=0 \implies yp(p^2+1) dp+(1-p^2)dy=0 ~~~~(3)$$
Comparing it with $M(p,y)dp+N(p,y)dy=0$, we get $\frac{\partial M}{\partial y}=p^3+p, \frac{\partial N}{\partial p}=-2p$ The integrating factor to make (3) exact is $$\mu(p)=\exp \int \frac{p^3+3p}{1-p^2} dp= \exp[-\frac{-p^2}{2}-2\ln(1-p^2)]=\frac{e^{-p^2/2}}{(1-p^2)^2}~~~(5)$$
Using this (3) becomes
$$yp(p^2+1)\frac{e^{-p^2/2}}{(1-p^2)^2} dp+ \frac{e^{-p^2/2}}{(1-p^2)} dy=0~~~~(6)$$
The solution of (6) can be written as
$$y \int p(p^2+1)\frac{e^{-p^2/2}}{(1-p^2)^2} dp =C$$ $$ \implies y= C e^{p^2/2}(1-p^2)~~~~(7)$$
Next using (1), we het
$$x=-Cpe^{p^2/2}~~~~(8)$$
Finally (7) and (8) are parametric solution of (1), where $p$ acts as a real parameter. In the following Fig. we plot (7,8) for $C=\pm 1,\pm 3/2, \pm 2$
as intersecting and self-orthogonal family of trajectories,

