Using Beltrami Identity on $ F=\frac{\sqrt{1+y'^2}}{y^2} $

Question

The problem is to make the following integral stationary: $$ \int_{x_1}^{x_2} \frac{\sqrt{1+y'^2}}{y^2}dx $$ since the functional doesn't depend explicitly on x, I decided to try Beltrami Identity: $$ F-y'\frac{\partial F}{\partial y'}=k $$ thus $$ \begin{aligned} \frac{\sqrt{1+y'^2}}{y^2}-y'\left(\frac{y'}{y^2\sqrt{1+y^2}}\right) &= k\\ y' &= \frac{\sqrt{1-k^2y^4}}{ky^2} \end{aligned} $$ and I get $$ \begin{aligned} y = \int \frac{\sqrt{1-k^2y^4}}{ky^2} dx &= \frac{\sqrt{1-k^2y^4}}{ky^2} \int dx\\ y &= \frac{x \sqrt{1-k^2y^4}}{ky^2}+C \end{aligned} $$ I think there's some errors somewhere, but I'm not sure. Is this answer correct?