Solve $$yy' = \sqrt{y^2+y'^2}y''-y'y''$$

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

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

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

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