Solution
= Solution
Use the <chain rule> along a sufficiently smooth <extremal> of the <functional>. For $E=y'F_{y'}-F$,
$$
\begin{aligned}
\frac{dE}{dx}
&=y''F_{y'}+y'\frac d{dx}F_{y'}
-F_x-F_yy'-F_{y'}y''\\
&=y'\left(\frac d{dx}F_{y'}-F_y\right)-F_x.
\end{aligned}
$$
The bracket vanishes by the <Euler-Lagrange equation>. When $F$ has no explicit $x$ dependence, $F_x=0$, giving the <Beltrami identity>:
$$
\boxed{y'F_{y'}-F=\text{constant along the extremal}.}
$$