Consider the following autonomous differential system: $$ x'=(y^2-1)(1-x^2-y^2)\\ y'=(x^2-1)(1-x^2-y^2) $$
We can easily see that $(\pm 1, \pm 1)$ are equilibria of the system.
We want to analyze its stability.
For the points $(1,1)$ and $(-1,-1)$ we have that the linealization of the system in those points has two non null eigenvalues (which are $\pm2$) so we know that the system is locally differentially equivalent to its linealization and we conclude that those points are saddle points.
However, the points $(1,-1)$ and $(-1,1)$ are trickier, since the eigenvalues there are purely complex (concretely $\pm 2 i$) so we do not have a guarantee of equivalence with the linealization.
How do I proceed?
As $$ (x^2-1)x'=(x^2-1)(y^2-1)(1-x^2-y^2)=(y^2-1)y' $$ one gets $$ E=\frac13x^3-x-\frac13y^3+y $$ as first integral. This function has saddle points at $(1,1)$ and $(-1,-1)$, a minimum at $(1,-1)$ and a maximum at $(-1,1)$. Around the extrema the level curves are concentric closed curves, and the solutions stay on the level set they started on.