Consider the nonlinear system:
$$\begin{cases}\dot{x}_1=(x_1-x_2)(1-x_1^2-x_2^2),\\\dot{x}_2=(x_1+x_2)(1-x_1^2-x_2^2).\end{cases}$$
Draw its phase portrait.
Solving $\dot{x}_1=\dot{x}_2=0$, we find equilibria $(0, 0)$ and $(\cos{\theta},\sin{\theta})$ for $0\le\theta<2\pi$. We can quickly find that $(0,0)$ is an unstable focus, however, how do we find the stability of the non-isolated equilibria $(\cos{\theta},\sin{\theta})$ for $0\le\theta<2\pi$?
Notice that $$ r'=\frac{x_1x_1'+x_2x_2'}{r}=r(1-r^2),\quad \theta'=\frac{x_1x_2'-x_2x_1'}{r^2}=1-r^2. $$ Hence, all equilibria on the unit circle are stable (but not asymptotically stable).
Added: If you look at the signs of $r'$ and $\theta'$ you will see that $r'>0$ for $r<1$ and that $r'<0$ for $r>0$. So all approaches $r=1$, except the origin. Moreover, since $\theta'$ has the same behavior, there are spirals inside and outside $r=1$ that go to an equilibrium point, rotating in the positive direction when $r<1$ and in the negative direction when $r>1$.
This information is sufficient to draw the whole phase portrait. From far away you will see: an equilibrium point at the origin, the equilibria on the unit circle, and then spirals that when passing through $r=1$ change the direction of motion (the last description is not rigorous but is much better than any rigorous description, since it really explains what is happening).