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.
 A: (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.
A: Your equations are $x˙=(x−2y)(x^2−1)$, $y˙=(3x+y)(x^2−1)$.  The equilibrium points are, as others have said, points where x'= 0 and y'= 0.  That is, $(x- 2y)(x^2- 1)$, $(3x+ y)(x^2- 1)= 0$.  First, if x= 1 or x= -1, y can be anything so every point on the line x= 1 or on the line x= -1 is an equilibrium point.  If $x^2- 1$ is not 0, then we must have x- 2y= 0 and 3x+ y= 0.  From the second equation, y= -3x so the first equation becomes x- 2(-3x)= 7x= 0, y= 0 and then x= 0.  (0, 0) is also an equilibrium point.  Whether that equilibrium point is "stable" or "unstable" depends on what happens close to (0, 0) so we can linearize the equations.  That is, if x is close to 0, $x^2$ is a lot closer- we can drop it and $y^2$.  The "linearized" equations are x'= x- 2y and y'= 3x+ y.  Differentiating again, x''= 1- 2y'= 1- 2(3x+ y)= 1- 6x+ 2y.  From the first equation 2y= x- x' so x''= 1- 6x+ x- x' or x''+ x'+ 5x= 1.  Solve that to find the trajectories and then use the original equation to see if solution move along those trajectories toward or away from (0, 0).
