= Solution
For an <equilibrium> at zero, the local <stable manifold> consists of nearby points whose forward trajectories remain nearby and tend to zero; the local <unstable manifold> is defined by backward trajectories with the analogous property. For a smooth system $\dot z=Az+f(z)$ with $f(0)=0$, $Df(0)=0$ and no zero-real-part <eigenvalues>, the stable-manifold theorem gives manifolds tangent to the negative- and positive-real-part eigenspaces. Their dimensions are the corresponding <eigenvalue> counts. Real distinct <eigenvalues> alone are insufficient if one is zero; then a <centre equilibrium> direction requires additional analysis. In the saddle case each manifold is a curve, found by imposing invariance on a power-series graph through the relevant <eigenvector>.
In this system, the $x$ <nullcline> is $y=x^2+2x$ and the $y$ <nullclines> are $x=0$ and $y=x-3$. Substituting the first into the second gives $x(x^2+x+3)=0$. Since the quadratic has <discriminant> $-11$, \b[the origin is the only real <equilibrium>]. Its <Jacobian matrix> $\left(\begin{smallmatrix}-2&1\\3&0\end{smallmatrix}\right)$ has <eigenvalues> $-3,1$, with <eigenvectors> $(1,-1)$ and $(1,3)$. It is therefore a hyperbolic saddle.
Set $y=h(x)$ for either manifold. Invariance requires
$$
h'(x)[-2x+h(x)-x^2]=3x+xh(x)-x^2.
$$
For $h=sx+bx^2+cx^3+O(x^4)$, matching the linear coefficient gives $s=-1$ or $s=3$. The quadratic and cubic coefficients respectively satisfy
$$
(3s-4)b=2s-1,\qquad(4s-6)c=b-2b(b-1).
$$
For the stable slope $s=-1$, this yields $b=3/7$, $c=-9/98$; for the unstable slope $s=3$, it gives $b=1$, $c=1/6$. Thus
$$
\boxed{W^s:\ y=-x+\frac37x^2-\frac9{98}x^3+O(x^4),\qquad W^u:\ y=3x+x^2+\frac16x^3+O(x^4).}
$$
Substituting back into $\dot x$ gives the reduced dynamics
$$
\boxed{\dot x|_{W^s}=-3x-\frac47x^2-\frac9{98}x^3+O(x^4),\qquad\dot x|_{W^u}=x+\frac16x^3+O(x^4).}
$$
The signs of the linear terms confirm attraction and repulsion along the respective curves.
In the <phase portrait>, $\dot x$ is positive above its parabolic <nullcline> and negative below it; $\dot y$ has sign $x(3+y-x)$. The stable branches approach the saddle and the unstable branches leave it in opposite directions. There can be no <periodic orbit>, since such an <orbit (dynamical system)> would require total enclosed <equilibrium> index $+1$, while the only <equilibrium> has index $-1$. There is no homoclinic excursion either: eliminating $y$ gives $\ddot x+(2+x)\dot x=x(x^2+x+3)$, so a positive turning point is a strict local minimum and a negative one a strict local maximum. A loop returning to zero would require the opposite extremum somewhere. Hence bounded forward trajectories lie on the <stable manifold> and approach zero; other trajectories eventually leave <compact> regions, without a claimed universal rate or infinite-time existence at infinity. The figure shows the global <nullclines> and the local cubic manifold approximations; the latter are used only near zero.
\Image[/past-exam-of-the-topics-in-applied-mathematics-course-of-the-university-of-oxford/2009-saddle-manifolds.png]
{title=The unique saddle, its nullclines and the local stable and unstable manifold approximations}
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