At a fixed point $x_0$, the gradient $\nabla H(x_0) = 0$. The Jacobian of the vector field $(H_y, -H_x)$ evaluated at $x_0$ is \begin{align*} A = \begin{pmatrix} H_{xy} & H_{yy} \\ -H_{xx} & -H_{xy} \end{pmatrix}. \end{align*} The trace satisfies $\operatorname{Tr} A = H_{xy} - H_{xy} = 0$. Since $\operatorname{Tr} A = \lambda_1 + \lambda_2 = 0$, the eigenvalues satisfy $\lambda_2 = -\lambda_1$. If $\lambda_1 \in \mathbb{R} \setminus \{0\}$, then $\lambda_2 = -\lambda_1$ has opposite sign, giving a saddle. If $\lambda_1 = i\omega$ with $\omega \neq 0$, then $\lambda_2 = -i\omega$, giving a centre. No focus is possible since $\operatorname{Re}(\lambda_1) = 0$ whenever $\operatorname{Tr} A = 0$. The identity $\dot{H} = H_x \dot{x} + H_y \dot{y} = H_x H_y - H_y H_x = 0$ shows $H$ is constant along trajectories.