= Solution
Solving the two factors in each equation gives \b[$(0,0),(3,0),(0,3),(1,1)$]. The Jacobian is
$$
Df=\begin{pmatrix}2x+2y-3&2x\\-2y&3-2x-2y\end{pmatrix}.
$$
At 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 $\left(\begin{smallmatrix}1&2\\-2&-1\end{smallmatrix}\right)$, with <eigenvalues> $\pm i\sqrt3$. The <linearization> is a centre, but the hyperbolic theorem does not apply there.
The nonlinear classification follows from the actual <Hamiltonian>
$$
\boxed{H(x,y)=xy(x+y-3),\qquad\dot x=H_y,\quad\dot y=-H_x.}
$$
Along solutions $\dot H=H_xH_y-H_yH_x=0$. At $(1,1)$, $H=-1$ and its <Hessian> is $\left(\begin{smallmatrix}2&1\\1&2\end{smallmatrix}\right)$, 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>. \b[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>.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2006/ii/paper-1-phase-plane.png]
{title=<Hamiltonian> level curves and flow around the three saddles and centre}
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