Solve $f'^2(x)(x^2 + f^2(x)) = s x (x f'^2(x) -x - 2 f(x) f'(x))$

Not elegant but it worked:

1. Put $f(x)=x h(x)$ then $$(x h'(x)+h(x))^2 (1 + h^2(x)) = s \left( x^2 h'^2(x)- h^2(x)-1\right).$$
2. Now, set $x=\exp(y)$ and $g = h\circ \exp$ then $$\left(g'(y)+g(y)\right)^2 \left(1 + g^2(y)\right) = s \left( g'^2(y)- g^2(y)-1\right).$$
3. If $g$ is invertible we can obtain for $y=g^{-1}(z)$ and $\phi=\frac{1}{(g^{-1})'}$ $$\left(\phi(z) +z\right)^2 \left(1 + z^2\right) = s \left( \phi^2(z)- z^2-1\right).$$
4. 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.