Solve $f'^2(x)(x^2 + f^2(x)) = s x (x f'^2(x) -x - 2 f(x) f'(x))$ Let $s>1$ and $f : \mathbb{R} \longrightarrow [0,\infty[$ differentiable be given.
How to solve
$$f'^2(x)(x^2  + f^2(x)) = s x (x f'^2(x) -x - 2 f(x) f'(x))\tag{1}$$
for $x$?
Is it even possible to find an explicit formula for $x$?
Using inverse function wasn't working for me.
This might be a particular example of the general problem
$$F(x,f(x),f'(x))=s.$$
Note that this is not a differential equation since $f$ is given.
Property: Let $s y$ solve (1) for $f$ then
$$f'^2(s y)(s^2 y^2  + f^2(s y)) = s^2 y (s y f'^2(s y) -s y - 2 f(s y) f'(s y)).$$
Using $h(y)=s^{-1}f(s y)$ gives
$$h'^2(y)(y^2  + h^2(y)) = s y (y h'^2(y) - y - 2 h(y) h'( y)).$$
So $y$ solves (1) for $h$.
 A: Not elegant but it worked:

*

*Put $f(x)=x h(x)$ then
$$(x h'(x)+h(x))^2 (1  + h^2(x)) = s \left( x^2 h'^2(x)- h^2(x)-1\right).$$

*Now, set $x=\exp(y)$ and $g = h\circ \exp$ then
$$\left(g'(y)+g(y)\right)^2 \left(1  + g^2(y)\right) = s \left( g'^2(y)- g^2(y)-1\right).$$

*If $g$ is invertible we can obtain for $y=g^{-1}(z)$ and $\phi=\frac{1}{(g^{-1})'}$
$$\left(\phi(z) +z\right)^2 \left(1  + z^2\right) = s \left( \phi^2(z)- z^2-1\right).$$

*Progressing with $\phi(z)=\frac{\sigma(z) \sqrt{1+z^2}-z(1+z^2)}{1+z^2-s} $ leads to
$$\sigma(z)^2= s^2-s,$$
with solution $z=\sigma^{-1}\left(\pm\sqrt{s^2-s}\right)$.

The solution to (1) is then
$$x= h^{-1} \left(\sigma^{-1}\left(\pm\sqrt{s^2-s}\right)\right)=[\sigma \circ h]^{-1}\left(\pm\sqrt{s^2-s}\right),$$
where $h(\xi)= \frac{f(\xi)}{\xi}$ and $\sigma$ reduces to
$$\sigma(z)= \frac{(1+z^2-s)f'(h^{-1}(z)) + s z}{\sqrt{1+z^2}},$$
and so
$$\sigma(h(\xi))= \frac{\xi(1+f^2(\xi)/\xi^2-s)f'(\xi) + s f(\xi)}{\sqrt{\xi^2+f^2(\xi)}}.$$
Note that due to the assumption $s>1$ a solution has to fulfill $2 |f(x)f'(x)|> |x|$.
I tried a symmetric non-trivial example for $f$ and it worked. The overall 6 solutions reduced to 4.
A: Not an answer, but there exists functions which produce any value of $x$ as a solution. For example, to get $x=\chi$ as a solution, consider
$$f(x)=x-\sqrt{s^2 \chi ^2-\chi ^2}-s \chi -\chi$$
Then $f'(\chi)=1$ and after a lot of simplification we get
$$\left(f(\chi)^2+\chi^2\right) f'(\chi)^2-s \chi \left(\chi f'(\chi)^2-2 f(\chi) f'(\chi)-\chi\right)=0$$
