$\omega$-limit set in $\mathbb S^2$ Let $X(x,y,z)=(-y+xz^2,x+yz^2,-zx^2-zy^2)$ be a vectorial field in $\mathbb S^2$, computationally it is easy to see that the $\omega-$limit set of any point other than the poles is the equator, but I don't know how to prove this mathematically, any ideas please?

 A: The poles are clearly fixed points. Consider any other point. To show that its omega limit set is equal to the equator, we calculate things explicitly.
First, let us verify that the omega limit set must be contained in the equator.
Since $x^2+y^2+z^2=1$ on the sphere, the $z$-direction of the vector field is $X_3(x,y,z) = -z(1-z^2)$.
Thus, the orbit $(x(t),y(t),z(t))$ satisfies $z'(t) = -z(t)(1-z(t)^2)$.
The solution to this ODE with initial value $z_0 \in (-1,1)$ is
$$
z(t) = \mathrm{sign}(z_0) \frac{1}{\sqrt{1+ce^{2t}}}
$$
with $c = z_0^{-2}-1$.
Since $c > 0$, we see from this that $z(t) \rightarrow 0$ as $t \rightarrow \infty$.
This shows that the orbit approaches the equator and, hence, the omega limit set must be a subset of the equator.
Second, we show that the omega limit set equals the whole equator.
This can be seen easiest in cylindrical coordinates.
In these, the vector field $X$ is expressed as
$$
X(r,\theta,z) =
\begin{pmatrix}
rz^2\\
-1\\
-zr^2
\end{pmatrix}.
$$
In particular, $\theta(t) = \theta(0)-t$ (mod $2\pi$), meaning that the orbit rotates counterclockwise around the equator with speed 1.
