Solve $$yy’ = sqrt{y^2+y’^2}y”-y’y”$$

First I set $p = y’$ and $p’ = frac{dp}{dy}p$ to form:

$$yp=sqrt{y^2+p^2}frac{dp}{dy}p-pfrac{dp}{dy}p rightarrow y=sqrt{y^2+p^2}frac{dp}{dy}-frac{dp}{dy}p$$

I am trying to come up with a clever substitution to deal with the square root and $p$ of :

$$y=frac{dp}{dy}(sqrt{y^2+p^2}-p)$$