[proofplan]
The proof is a chain-rule computation for the Legendre transform. First we differentiate the Hamiltonian $H(x,y,p)=p\cdot v(x,y,p)-L(x,y,v(x,y,p))$ and use the defining relation $p=\partial_vL(x,y,v)$ to cancel all terms involving derivatives of the inverse velocity map. The two resulting identities, $\partial_pH=v$ and $\partial_yH=-\partial_yL$, convert the Euler-Lagrange equation into Hamilton's equations and convert Hamilton's equations back into the Euler-Lagrange equation.
[/proofplan]
[step:Derive the differential identities for the Hamiltonian]
Let
\begin{align*}
V:W\to U
\end{align*}
denote the map
\begin{align*}
V(x,y,p):=(x,y,v(x,y,p)).
\end{align*}
Since $W$ is relatively open by hypothesis, and since $L\in C^2(U;\mathbb R)$ and $v\in C^1(W;\mathbb R^n)$, the Hamiltonian $H\in C^1(W;\mathbb R)$. Thus the following partial derivatives are taken on the relatively open domain $W$. Fix $(x,y,p)\in W$, and put
\begin{align*}
\tilde v:=v(x,y,p).
\end{align*}
By the defining property of the inverse Legendre map,
\begin{align*}
p=\partial_vL(x,y,\tilde v).
\end{align*}
For each $j\in\{1,\dots,n\}$, the chain rule gives
\begin{align*}
\frac{\partial H}{\partial p_j}(x,y,p)=\tilde v_j+\sum_{i=1}^n p_i\frac{\partial v_i}{\partial p_j}(x,y,p)-\sum_{i=1}^n \frac{\partial L}{\partial v_i}(x,y,\tilde v)\frac{\partial v_i}{\partial p_j}(x,y,p).
\end{align*}
Using $p_i=\frac{\partial L}{\partial v_i}(x,y,\tilde v)$ for every $i$, the two sums cancel, so
\begin{align*}
\frac{\partial H}{\partial p_j}(x,y,p)=\tilde v_j.
\end{align*}
Since this holds for each component $j$,
\begin{align*}
\partial_pH(x,y,p)=v(x,y,p).
\end{align*}
For each $j\in\{1,\dots,n\}$, the chain rule with respect to the coordinate $y_j$ gives
\begin{align*}
\frac{\partial H}{\partial y_j}(x,y,p)=\sum_{i=1}^n p_i\frac{\partial v_i}{\partial y_j}(x,y,p)-\frac{\partial L}{\partial y_j}(x,y,\tilde v)-\sum_{i=1}^n \frac{\partial L}{\partial v_i}(x,y,\tilde v)\frac{\partial v_i}{\partial y_j}(x,y,p).
\end{align*}
Again using $p_i=\frac{\partial L}{\partial v_i}(x,y,\tilde v)$, the two sums cancel, and therefore
\begin{align*}
\frac{\partial H}{\partial y_j}(x,y,p)=-\frac{\partial L}{\partial y_j}(x,y,\tilde v).
\end{align*}
Thus
\begin{align*}
\partial_yH(x,y,p)=-\partial_yL(x,y,v(x,y,p)).
\end{align*}
[guided]
The Hamiltonian depends on $p$ both directly through the dot product $p\cdot v(x,y,p)$ and indirectly through the velocity $v(x,y,p)$. The point of the Legendre transform is that the indirect terms cancel because $p$ is exactly the velocity gradient of $L$.
Define the map
\begin{align*}
V:W\to U
\end{align*}
by
\begin{align*}
V(x,y,p):=(x,y,v(x,y,p)).
\end{align*}
Because $W$ is relatively open by hypothesis, the partial derivatives of $H$ are taken on a genuine relative domain. The assumptions $L\in C^2(U;\mathbb R)$ and $v\in C^1(W;\mathbb R^n)$ imply, by the chain rule for $C^1$ maps, that $H\in C^1(W;\mathbb R)$. Fix $(x,y,p)\in W$ and define
\begin{align*}
\tilde v:=v(x,y,p).
\end{align*}
The inverse Legendre relation says precisely that
\begin{align*}
p=\partial_vL(x,y,\tilde v).
\end{align*}
We first differentiate $H$ with respect to $p_j$, where $j\in\{1,\dots,n\}$. Since
\begin{align*}
H(x,y,p)=p\cdot v(x,y,p)-L(x,y,v(x,y,p)),
\end{align*}
the product rule applied to the dot product gives one direct term $\tilde v_j$ and one indirect term involving $\partial v_i/\partial p_j$. The chain rule applied to $L(x,y,v(x,y,p))$ gives another indirect term involving the same derivatives of $v$. Therefore
\begin{align*}
\frac{\partial H}{\partial p_j}(x,y,p)=\tilde v_j+\sum_{i=1}^n p_i\frac{\partial v_i}{\partial p_j}(x,y,p)-\sum_{i=1}^n \frac{\partial L}{\partial v_i}(x,y,\tilde v)\frac{\partial v_i}{\partial p_j}(x,y,p).
\end{align*}
Now substitute the defining identity $p_i=\frac{\partial L}{\partial v_i}(x,y,\tilde v)$. The two sums are identical with opposite signs, hence
\begin{align*}
\frac{\partial H}{\partial p_j}(x,y,p)=\tilde v_j.
\end{align*}
Since this holds for every component $j$, we obtain the vector identity
\begin{align*}
\partial_pH(x,y,p)=v(x,y,p).
\end{align*}
The derivative with respect to $y$ is similar, except there is no direct derivative of $p$ because $p$ is held fixed when taking the partial derivative $\partial_yH$. For $j\in\{1,\dots,n\}$, the chain rule gives
\begin{align*}
\frac{\partial H}{\partial y_j}(x,y,p)=\sum_{i=1}^n p_i\frac{\partial v_i}{\partial y_j}(x,y,p)-\frac{\partial L}{\partial y_j}(x,y,\tilde v)-\sum_{i=1}^n \frac{\partial L}{\partial v_i}(x,y,\tilde v)\frac{\partial v_i}{\partial y_j}(x,y,p).
\end{align*}
Again the Legendre relation $p_i=\frac{\partial L}{\partial v_i}(x,y,\tilde v)$ cancels the two sums. What remains is
\begin{align*}
\frac{\partial H}{\partial y_j}(x,y,p)=-\frac{\partial L}{\partial y_j}(x,y,\tilde v).
\end{align*}
Since this holds for every component $j$,
\begin{align*}
\partial_yH(x,y,p)=-\partial_yL(x,y,v(x,y,p)).
\end{align*}
These two identities are the only special facts about the Legendre transform needed in the rest of the proof.
[/guided]
[/step]
[step:Convert the Euler-Lagrange equation into Hamilton's equations]
Assume $y\in C^2([a,b];\mathbb R^n)$, assume $(x,y(x),y'(x))\in U$ for every $x\in[a,b]$, and define
\begin{align*}
p:[a,b]\to\mathbb R^n
\end{align*}
by
\begin{align*}
p(x):=\partial_vL(x,y(x),y'(x)).
\end{align*}
Since $L\in C^2(U;\mathbb R)$ and $y\in C^2([a,b];\mathbb R^n)$, the curve $p$ is of class $C^1$. By the definition of $W$, the point $(x,y(x),p(x))$ belongs to $W$ for every $x\in[a,b]$.
The inverse Legendre relation gives
\begin{align*}
v(x,y(x),p(x))=y'(x).
\end{align*}
Using $\partial_pH=v$ from the previous step, we obtain
\begin{align*}
y'(x)=\partial_pH(x,y(x),p(x)).
\end{align*}
This is the first Hamilton equation.
The Euler-Lagrange equation states
\begin{align*}
\partial_yL(x,y(x),y'(x))-\frac{d}{dx}\partial_vL(x,y(x),y'(x))=0.
\end{align*}
Since $p(x)=\partial_vL(x,y(x),y'(x))$, this is equivalent to
\begin{align*}
p'(x)=\partial_yL(x,y(x),y'(x)).
\end{align*}
Using $y'(x)=v(x,y(x),p(x))$ and $\partial_yH=-\partial_yL(\cdot,\cdot,v(\cdot,\cdot,\cdot))$ from the previous step, we get
\begin{align*}
p'(x)=-\partial_yH(x,y(x),p(x)).
\end{align*}
This is the second Hamilton equation.
[/step]
[step:Convert Hamilton's equations back into the Euler-Lagrange equation]
Conversely, assume $y,p\in C^1([a,b];\mathbb R^n)$, assume $(x,y(x),p(x))\in W$ for all $x\in[a,b]$, and assume Hamilton's equations hold. The first Hamilton equation and the identity $\partial_pH=v$ give
\begin{align*}
y'(x)=v(x,y(x),p(x)).
\end{align*}
Since $v\in C^1(W;\mathbb R^n)$, $y,p\in C^1([a,b];\mathbb R^n)$, and $(x,y(x),p(x))\in W$ for all $x\in[a,b]$, the composition $x\mapsto v(x,y(x),p(x))$ is of class $C^1$. Hence $y'\in C^1([a,b];\mathbb R^n)$, so $y\in C^2([a,b];\mathbb R^n)$. Therefore $(x,y(x),y'(x))\in U$ and, by the defining property of the inverse Legendre map,
\begin{align*}
p(x)=\partial_vL(x,y(x),y'(x)).
\end{align*}
Because $p\in C^1([a,b];\mathbb R^n)$, we may differentiate this identity along the curve and obtain
\begin{align*}
p'(x)=\frac{d}{dx}\partial_vL(x,y(x),y'(x)).
\end{align*}
The second Hamilton equation and the identity $\partial_yH=-\partial_yL(\cdot,\cdot,v(\cdot,\cdot,\cdot))$ give
\begin{align*}
p'(x)=-\partial_yH(x,y(x),p(x))=\partial_yL(x,y(x),v(x,y(x),p(x))).
\end{align*}
Using $v(x,y(x),p(x))=y'(x)$, this becomes
\begin{align*}
p'(x)=\partial_yL(x,y(x),y'(x)).
\end{align*}
Combining the two displayed identities for $p'(x)$ yields
\begin{align*}
\frac{d}{dx}\partial_vL(x,y(x),y'(x))=\partial_yL(x,y(x),y'(x)).
\end{align*}
Equivalently,
\begin{align*}
\partial_yL(x,y(x),y'(x))-\frac{d}{dx}\partial_vL(x,y(x),y'(x))=0.
\end{align*}
Thus $y$ satisfies the Euler-Lagrange equation, and the equivalence is proved.
[/step]
