 Solution

The equilibria are $\boxed{(0,0),(\pm \sqrt{2/3},0)}$. The Jacobian is $J=\left(\begin{array}{cc}0& 1\\ 2-9x^2& -1\end{array}\right)$. At the origin it has eigenvalues $1,-2$ and tangent eigenvectors $(1,1)$ and $(1,-2)$, so the origin is a saddle equilibrium. At either nonzero point the eigenvalues are $\boxed{(-1\pm i\sqrt{15})/2}$, giving stable foci.

The energy-like function $H=y^2/2-x^2+3x^4/4$ has $\dot{H}=-y^2\le 0$. It is coercive at large distance, so trajectories remain bounded; there are no periodic orbits because their energy could not return to its starting value unless $y\equiv 0$. The stable manifold of the saddle equilibrium separates the two attracting basins. Its unstable branches leave the origin and approach the right and left foci, respectively. Their negative energy near the origin prevents crossing back through $x=0$.

For either invariant manifold write $y=h(x)=sx+c{x}^3+O(x^5)$. Odd symmetry rules out even powers. Tangency to the vector field requires $h^{\prime }(x)h(x)=2x-h(x)-3x^3$. Matching coefficients yields $s^2+s-2=0$ and $c(4s+1)=-3$. HenceThese are local graphs, not globally exact cubic curves.