= Solution
At a hyperbolic saddle, the stable invariant subspace of $Df(0)$ is the sum of its <generalized eigenspaces> with negative real parts; the unstable subspace uses positive real parts. The <stable manifold theorem> gives invariant local manifolds tangent to these subspaces at the equilibrium, with the corresponding dimensions and the appropriate forward or backward convergence to the equilibrium.
Here the Jacobian is $\operatorname{diag}(1,-1)$, so the unstable tangent is the $x$ axis and the stable tangent the $y$ axis. For the stable graph write $x=h(y)=Ay^2+By^3+O(y^4)$. Invariance requires $h'(y)(-y+3h(y)^2)=h+h^2+2hy+3y^2$. Comparing quadratic and cubic terms gives $-2A=A+3$ and $-3B=B+2A$, so $A=-1,B=1/2$.
For the unstable graph $y=g(x)=Cx^2+Dx^3+O(x^4)$, invariance is $g'(x)(x+x^2+2xg+3g^2)=-g+3x^2$. Coefficients give $2C=3-C$ and $2C+3D=-D$, so $C=1,D=-1/2$. Thus the <cubic graph expansion of a saddle invariant manifold> is
$$
\boxed{W^s:\ x=-y^2+y^3/2+O(y^4),\qquad W^u:\ y=x^2-x^3/2+O(x^4).}
$$
These describe the local manifolds to cubic order, not exact global <polynomial> curves.
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