[proofplan] We prove the result by differentiating the energy function $H\circ\gamma$ along the integral curve. The integral-curve equation replaces $\dot{\gamma}(t)$ by $X_H(\gamma(t))$, and the defining identity for the Hamiltonian vector field converts this derivative into $\omega(X_H,X_H)$. Since every two-form is alternating, this expression vanishes at each time, so the derivative of $H\circ\gamma$ is identically zero. [/proofplan] [step:Differentiate the Hamiltonian along the integral curve] Define the smooth function $E:I\to\mathbb{R}$ by \begin{align*} E(t)=H(\gamma(t)). \end{align*} For each $t\in I$, differentiating the composite $H\circ\gamma$ gives \begin{align*} E'(t)=dH_{\gamma(t)}(\dot{\gamma}(t)). \end{align*} Since $\gamma$ is an integral curve of $X_H$, we have $\dot{\gamma}(t)=X_H(\gamma(t))$. Therefore \begin{align*} E'(t)=dH_{\gamma(t)}(X_H(\gamma(t))). \end{align*} [guided] We introduce the energy along the trajectory as the map $E:I\to\mathbb{R}$ defined by \begin{align*} E(t)=H(\gamma(t)). \end{align*} The goal is to show that this real-valued function is constant. Since $E$ is smooth, it is enough to compute its derivative at an arbitrary time $t\in I$. Differentiating the composite map $H\circ\gamma$ gives \begin{align*} E'(t)=dH_{\gamma(t)}(\dot{\gamma}(t)). \end{align*} Here $dH_{\gamma(t)}:T_{\gamma(t)}M\to\mathbb{R}$ is the differential of $H$ at the point $\gamma(t)$, and $\dot{\gamma}(t)\in T_{\gamma(t)}M$ is the velocity vector of the curve at time $t$. Now we use the defining property of an integral curve. The hypothesis says precisely that the velocity of $\gamma$ at time $t$ is the Hamiltonian vector field evaluated at the point $\gamma(t)$: \begin{align*} \dot{\gamma}(t)=X_H(\gamma(t)). \end{align*} Substituting this identity into the derivative formula gives \begin{align*} E'(t)=dH_{\gamma(t)}(X_H(\gamma(t))). \end{align*} This reduces conservation of $H$ to a pointwise algebraic identity involving the symplectic form. [/guided] [/step] [step:Use alternation of the symplectic form to make the derivative vanish] Fix $t\in I$ and set $p=\gamma(t)$. By the defining identity for $X_H$, applied at the point $p$ with the tangent vector $v=X_H(p)$, we obtain \begin{align*} dH_p(X_H(p))=\omega_p(X_H(p),X_H(p)). \end{align*} Because $\omega_p:T_pM\times T_pM\to\mathbb{R}$ is alternating, every vector pairs with itself to zero. Hence \begin{align*} \omega_p(X_H(p),X_H(p))=0. \end{align*} Combining this with the derivative formula from the previous step gives \begin{align*} E'(t)=0. \end{align*} Since $t\in I$ was arbitrary, $E'=0$ on $I$. [/step] [step:Conclude that the Hamiltonian is constant along the trajectory] The function $E:I\to\mathbb{R}$ is smooth and satisfies $E'(t)=0$ for every $t\in I$. Therefore $E$ is constant on the interval $I$. Since $E=H\circ\gamma$, this means that $H(\gamma(t))$ is constant in $t$, as claimed. [/step]