= Solution
Define $h=f-y'f_{y'}$. For a density without explicit $x$ dependence, the chain rule gives
$$
\frac{dh}{dx}=f_yy'+f_{y'}y''-y''f_{y'}-y'\frac d{dx}f_{y'}
=y'\left(f_y-\frac d{dx}f_{y'}\right).
$$
Thus the <Euler-Lagrange equation> implies the <Beltrami identity>
$$
\boxed{h=f-y'\frac{\partial f}{\partial y'}=\text{constant},\qquad\frac{dh}{dx}=0.}
$$
Where $y'\ne0$ this first integral is equivalent to the original equation. At a turning point it extends by continuity. On an entire constant segment, however, its derivative alone can vanish without imposing the full <Euler-Lagrange equation>, so such candidate paths must still be checked in the original equation.
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