This has been solved :was an overwrite error
substitution in beltrami identity: $$F(y,y')= \sqrt{\frac{1+y'^2}{2gy}}$$ $$\frac{\partial F}{\partial y'} = \frac{y'^2}{\sqrt{1+y'^2}\sqrt{2gy}}$$
Beltrami identity:$$F-y'\frac{\partial F}{\partial y'} = C$$
My answer: $\sqrt{\frac{1+y'^2 }{2gy}} -\frac{y'^2}{\sqrt{1+y'^2}\sqrt{2gy}}=C$ simplifying gives: $\frac{1}{\sqrt{y(y'^2+1)}}=C\sqrt{2g}$
final simplifying answer :
$$y(1+y'^2)=\frac{1}{2gC^2}=2R$$
i am clearly substituting wrong, can someone help? has been resolved thx :)