Consider the system $\dot{x}=(x-2y)(x^2-1), \dot{y}=(3x+y)(x^2-1)$ Consider the system 
$$\dot{x}=(x-2y)(x^2-1), \dot{y}=(3x+y)(x^2-1)$$
Determine the stability property of the system equilibria. If an equilibrium is asymptotically stable, give an estimate of its attractiveness basin. (For $(0,0)$ use the Liaponouv $V(x,y)=3x^2+2y^2$ function)
I have already tried to solve this problem and I get to that $(0,0)$ is a balance point but I do not know how to find the others, is there only this balance point? I already found the attractiveness basin of $(0,0)$, but I need to know the other points of balance.

(x, y) is an "equilibrium point" (your "balance point" though it seems strange that you would ask about "system equilibria" without using the phrase "equilibrium point") if and only if $x'= (x- 2y)(x^2- 1)= 0$ and $y'= (3x+ y)(x^2- 1)= 0$.  That will clearly be true if $x^2- 1= 0$.  That is, if x= 1 or -1 with y anything.  Every point along the lines x= 1 and x= -1 is an equilibrium point.  If $x^2- 1\ne 0$, we must have both x- 2y= 0 and 3x+ y= 0 and, of course, the only (x, y) that satisfies those is (0, 0).  That is, the equilibrium points are (0, 0), (1, y) for all y, and (-1, y) for all y.
