Take a look at the notes on lectures 1 and 2 of Geometric Numerical Integration found here. Quoting from Lecture 2

A numerical one-step method $y_{n+1} = \Phi_h(y_n)$ is called symplectic if, when applied
to a Hamiltonian system, the discrete flow $y \mapsto \Phi_h(y)$ is a symplectic transformation for all sufficiently small step sizes.

From your link you have
$$x(t+h) = x(t) + h \dot{x}(t) + h^2 \left\{\frac{3}{24}f(t+h) +\frac{10}{24}f(t) -\frac{1}{24}f(t-h) \right\}$$
and
$$\dot{x}(t+h) = \frac{x(t+h) - x(t)}{h} + h \dot{x}(t) + h \left\{\frac{7}{24}f(t+h) +\frac{6}{24}f(t) -\frac{1}{24}f(t-h) \right\}$$

Now take $\omega(\xi,\eta) = \xi^T J \eta$ where
$J = \left(\begin{array}{cc} 0 & \mathbb{I} \\ \mathbb{I} & 0 \end{array}\right)$. Then the integrator is symplectic if and only if $\omega(x(t),\dot{x}(t))=\omega(x(t+h),\dot{x}(t+h))$ for sufficiently small $h$.

All that you need to do is to fill in the values of $x(t+h)$ and $\dot{x}(t+h)$ from the integrator, and show that this condition does not hold.

This post has been migrated from (A51.SE)