Region of attraction and asymptotics stable points-Liapunov's Function (1/2) PROBLEM:
1)Show that the stationary point O(0,0) is asymptotic stable
2)Find a region of attraction for the system
Our system is: $$ x\cdot (x-a)=x' $$
$$y\cdot (y-b)=y' $$
a,b>0
given the liapunov's function $$V=(\dfrac{x}{a})^2+(\dfrac{y}{b})^2 $$
So,i differentiate my liapunov's function and i take:
$$ V'=2\cdot x^2 \cdot \dfrac{x-a}{a} + 2 \cdot y^2 \cdot \dfrac{y-b}{b} $$
so,now i have to show that this is negative  but i didn't reach to a conclusion for the sign of V'.
For question 2 and the region of attraction ,i am not sure What i have to do.To be honest the examples of my book of dynamical's system don't help a lot.It's the first time i meet a problem of this kind so,i haven't cleared my mind on this.
I would really appreciate a thorough solution and explanation about how to find a region of attraction.Also,i would really appreciate if someone can show me a way to show that V' is negative  as we want O(0,0) to be an asymptotic stable point.
Thanks in advance!
 A: You should always write your differential equations from left to right. 
$$\dot{x}=-ax+x^2$$
$$\dot{y}=-by+y^2$$
Note, that the asymptotic stability of the origin directly follows from a linearization and $a>0$ and $b>0$. But we can also use your Lyapunov function $V(x,y)=\left(\dfrac{x}{a}\right)^2+\left(\dfrac{y}{b}\right)^2$. The time derivative of $V$ is given as
$$\dot{V}=-2x^2\left(1-x/a\right)-2y^2(1-y/b).$$
If we assume $a>0$ and $b>0$ we can see that $\dot{V}$ is negative definite if $1-x/a>0 \implies a>x$ and $1-y/b>0 \implies b>y$ is given. 
In order to get the region of attraction, we need to consider level sets of $V< c$. We can directly see that these level sets are ellipses with $c=1$. We have to choose $c=1$ or we would not be able to fulfil $a>x$ and $b>y$ Inside these ellipses $\dot{V}$ is negative definite. Hence, the boundary of the estimation of the region of attraction is given by $$(x/a)^2+(y/b)^2=1.$$
A: *

*To show asymptotical stability you could use linearization:
$J(x,y) = \begin{bmatrix}2x - a & 0\\0 & 2y - b\end{bmatrix}$
That in $O=(0,0)$:
$J(0,0) = \begin{bmatrix} - a & 0\\0 & - b\end{bmatrix}$
So you can easily see that the eigenvalues are exactly $\lambda_{1}=
-a$ and $\lambda_{2} = -b$, since $a,b>0$ you have asymptotic stability for $(0,0)$

*The Lyapunov function that you wrote is ok and so you can see that
$V'<0$ if for example $x<a$ and $y<b$. So a possible field of
attraction (using La Salle) could be the region:
$B= \{ (x,y): x^{2} + y^{2} < m \}$   where $m = min(a,b)$
