[proofplan]
We expand the Euler-Lagrange equation along the curve $x \mapsto (x,y(x),y'(x))$ using the chain rule applied to the map $\partial_v L$. This produces an equation that is linear in $y''(x)$, whose coefficient is $\partial_{vv}L(x,y(x),y'(x))$. Regularity and continuity make this coefficient nonzero on a small relative neighbourhood of the chosen point, so division defines a [continuous function](/page/Continuous%20Function) $F$ and converts the Euler-Lagrange equation into the asserted second-order ODE form.
[/proofplan]
[step:Choose a neighbourhood where the regularity coefficient remains nonzero]
Fix $x_0 \in [a,b]$, and define
\begin{align*}
A=[a,b]\times\mathbb{R}\times\mathbb{R}.
\end{align*}
Since $L \in C^2(A;\mathbb{R})$, the second partial derivative
\begin{align*}
G: A \to \mathbb{R}, \quad G(x,u,v)=\partial_{vv}L(x,u,v)
\end{align*}
is continuous. By regularity along $y$,
\begin{align*}
G(x_0,y(x_0),y'(x_0))=\partial_{vv}L(x_0,y(x_0),y'(x_0)) \neq 0.
\end{align*}
Because $G$ is continuous, there exists a relative neighbourhood $U$ of $(x_0,y(x_0),y'(x_0))$ in $A$ such that
\begin{align*}
G(x,u,v)\neq 0
\end{align*}
for every $(x,u,v)\in U$. Equivalently, $\partial_{vv}L\neq 0$ on $U$.
[guided]
Fix a point $x_0 \in [a,b]$. The coefficient that will later multiply $y''$ is the second derivative of $L$ with respect to its velocity variable. To isolate this coefficient, define the ambient set
\begin{align*}
A=[a,b]\times\mathbb{R}\times\mathbb{R}
\end{align*}
and define the function
\begin{align*}
G: A \to \mathbb{R}, \quad G(x,u,v)=\partial_{vv}L(x,u,v).
\end{align*}
This is a genuine continuous function because $L \in C^2(A;\mathbb{R})$. The regularity hypothesis says exactly that this function does not vanish at every point of the lifted curve $x \mapsto (x,y(x),y'(x))$. In particular,
\begin{align*}
G(x_0,y(x_0),y'(x_0))\neq 0.
\end{align*}
A continuous real-valued function that is nonzero at a point remains nonzero on some neighbourhood of that point. Since $x_0$ may be an endpoint of $[a,b]$, the neighbourhood is taken in the relative topology of $A=[a,b]\times\mathbb{R}\times\mathbb{R}$. Hence there is a relative neighbourhood $U$ of $(x_0,y(x_0),y'(x_0))$ such that
\begin{align*}
\partial_{vv}L(x,u,v)=G(x,u,v)\neq 0
\end{align*}
for every $(x,u,v)\in U$. This is the local nonvanishing condition needed for division by $\partial_{vv}L$.
[/guided]
[/step]
[step:Expand the Euler-Lagrange equation along the lifted curve]
Define the lifted curve
\begin{align*}
\gamma: [a,b]\to A, \quad \gamma(x)=(x,y(x),y'(x)).
\end{align*}
Since $y\in C^2([a,b];\mathbb{R})$ and $\partial_v L\in C^1(A;\mathbb{R})$, the composite map
\begin{align*}
H: [a,b]\to\mathbb{R}, \quad H(x)=\partial_v L(\gamma(x))
\end{align*}
is differentiable on $(a,b)$. By the chain rule, for every $x\in(a,b)$,
\begin{align*}
H'(x)=\partial_{xv}L(x,y(x),y'(x))+\partial_{uv}L(x,y(x),y'(x))\,y'(x)+\partial_{vv}L(x,y(x),y'(x))\,y''(x).
\end{align*}
Substituting this identity into the Euler-Lagrange equation gives
\begin{align*}
\partial_u L(x,y(x),y'(x))-\partial_{xv}L(x,y(x),y'(x))-\partial_{uv}L(x,y(x),y'(x))\,y'(x)-\partial_{vv}L(x,y(x),y'(x))\,y''(x)=0
\end{align*}
for every $x\in(a,b)$.
[/step]
[step:Solve the local linear equation for $y''$]
Let $x\in(a,b)$ satisfy $\gamma(x)\in U$. Since $\partial_{vv}L\neq 0$ on $U$, the coefficient
\begin{align*}
\partial_{vv}L(x,y(x),y'(x))
\end{align*}
is nonzero. Rearranging the equation obtained above and dividing by this nonzero coefficient gives
\begin{align*}
y''(x)=\frac{\partial_u L(x,y(x),y'(x))-\partial_{xv}L(x,y(x),y'(x))-y'(x)\,\partial_{uv}L(x,y(x),y'(x))}{\partial_{vv}L(x,y(x),y'(x))}.
\end{align*}
Thus the Euler-Lagrange equation is a second-order equation solved explicitly for $y''$ wherever the lifted curve lies in $U$.
[/step]
[step:Define the continuous local right hand side]
Define
\begin{align*}
F: U\to\mathbb{R}, \quad F(x,u,v)=\frac{\partial_u L(x,u,v)-\partial_{xv}L(x,u,v)-v\,\partial_{uv}L(x,u,v)}{\partial_{vv}L(x,u,v)}.
\end{align*}
The functions $\partial_u L$, $\partial_{xv}L$, $\partial_{uv}L$, and $\partial_{vv}L$ are continuous on $A$ because $L\in C^2(A;\mathbb{R})$. The denominator $\partial_{vv}L$ is nonzero on $U$ by construction, so $F$ is continuous on $U$ as a quotient of continuous functions with nonvanishing denominator. The identity from the previous step is precisely
\begin{align*}
y''(x)=F(x,y(x),y'(x))
\end{align*}
for every $x\in(a,b)$ with $(x,y(x),y'(x))\in U$. This proves the claimed local ODE form.
[/step]
