The Beltrami identity

Define $H(x) = L(y,y') - y'\,\frac{\partial L}{\partial y'}$ evaluated along an extremal $y(x)$, and differentiate with respect to $x$ using the chain rule: $\frac{dH}{dx} = \frac{\partial L}{\partial y} y' + \frac{\partial L}{\partial y'} y'' - y''\,\frac{\partial L}{\partial y'} - y'\,\frac{d}{dx}\frac{\partial L}{\partial y'}$.

The two terms containing $y''$ cancel exactly, leaving $\frac{dH}{dx} = y'\left( \frac{\partial L}{\partial y} - \frac{d}{dx}\frac{\partial L}{\partial y'} \right)$.

But $y(x)$ is an extremal, so by the Euler–Lagrange equation the bracketed factor is identically zero along $y$. Hence $\frac{dH}{dx}=0$ everywhere on the interval, which means $H(x)=L-y'\,\frac{\partial L}{\partial y'}$ is constant, proving the claim.