The Beltrami Identity:
If the Lagrangian $\:L\left(y,y',x\right)\:$ of a system does not depend explicitly on $\:x$, that is \begin{equation} \dfrac{\partial L}{\partial x}=0 \tag{01}\label{01} \end{equation} then from the Euler-Lagrange equation \begin{equation} \dfrac{\mathrm{d}}{\mathrm{d}x}\left(\dfrac{\partial L}{\partial y'}\right)-\dfrac{\partial L}{\partial y}=0 \tag{02}\label{02} \end{equation} we have \begin{equation} \dfrac{\mathrm{d}}{\mathrm{d}x}\left(y'\dfrac{\partial L}{\partial y'}-L\right)=0 \tag{03}\label{03} \end{equation} so \begin{equation} \boxed{\:\:y'\dfrac{\partial L}{\partial y'}-L = \texttt{constant}\:\:}\quad \texttt{(Beltrami Identity)} \tag{04}\label{04} \end{equation} For your Lagrangian \begin{equation} \begin{split} \frac{\sqrt{\dot x ^2 + \dot y ^2}}{y}\mathrm dt &= \frac{\sqrt{1 +\left(\dfrac{\dot y}{\dot x}\right)^2}}{y}\dot x\,\mathrm dt =\frac{\sqrt{1 +\left(\dfrac{\mathrm dy/\mathrm dt}{\mathrm dx/\mathrm dt}\right)^2}}{y}\dfrac{\mathrm dx}{\mathrm dt}\,\mathrm dt\\ &=\frac{\sqrt{1 +\left(\dfrac{\mathrm dy}{\mathrm dx}\right)^2}}{y}\mathrm dx=\frac{\sqrt{1 + y'^{2}}}{y}\mathrm dx\\ \end{split} \tag{05}\label{05} \end{equation} that is \begin{equation} L\left(y,y',x\right)=\frac{\sqrt{1 + y'^{2}}}{y} \tag{06}\label{06} \end{equation}