# Show that zero attractor basin is not $\mathbb{R}^2$

I have a dynamical system $$\left\{ \begin{array}{rcl} \dot {x_1} & = & -\frac{2x_{1}}{(1+x_{1}^2)^2} - \frac{2x_2}{(1+x_2^2)^2} \\ \dot {x_2} & = & 2x_1 - \frac{2 x_2}{(1+x_2^2)^2} \end{array} \right.$$ Here $(x_1, x_2) \in \mathbb{R}^2$. Zero solution is an attractor, but I know that zero solution is not a global attractor. How to show this analitically (without writing a program)? Any ideas are welcome.

-
Why is it not a global attractor? How do you know that? – Pantelis Sopasakis Dec 4 '12 at 2:28
@PantelisSopasakis I wrote a program. Furthermore, it is a homework problem! – Nimza Dec 4 '12 at 7:14