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Solving the two factors in each equation gives $(0,0),(3,0),(0,3),(1,1)$. The Jacobian isAt the three boundary equilibria its eigenvalues are $3,-3$. These equilibria are hyperbolic fixed points; the Hartman-Grobman theorem gives local topological conjugacy to their saddle linearizations. At $(1,1)$ the matrix is $\left(\begin{array}{cc}1& 2\\ -2& -1\end{array}\right)$, with eigenvalues $\pm i\sqrt{3}$. The linearization is a centre, but the hyperbolic theorem does not apply there.

The nonlinear classification follows from the actual HamiltonianAlong solutions $\dot{H}=H_x{H}_y-H_y{H}_x=0$. At $(1,1)$, $H=-1$ and its Hessian is $\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$, which is positive definite. Nearby regular level curves are closed ovals around this strict minimum. Their vector field is nonzero and tangent, so each is a periodic orbit. The nonlinear equilibrium is a centre.

The separatrix level $H=0$ consists of $x=0$, $y=0$, and $x+y=3$. Within the triangle they connect the saddles as $(3,0)\to (0,0)\to (0,3)\to (3,0)$. The interior levels $-1<H<0$ are nested clockwise periodic orbits; outside the triangle the remaining level branches are unbounded. On $y=0$, flow is left for $0<x<3$ and right for $x<0$ or $x>3$; on $x=0$, it is up for $0<y<3$ and down otherwise. These directions and the invariant levels determine the full phase portrait.