Stability of Critical Points in a System of DEs

[fluidistic]

Homework Statement

Find the stationary solution(s) of the following system of DE and determine its stability:
$x'=x-x^3-xy^2$.
$y'=2y-y^5-x^4y$.

Homework Equations

$x'=0, y'=0$.
Expansion of x and y around the critical point: $x=x_0+\alpha$, $y=y_0+\beta$.

The Attempt at a Solution

I solved $x'=y'=0$, this gave me $x= \pm \sqrt {1-y^2}$ or $x=0$. And $y=\pm \sqrt {\pm \sqrt {2-x^4}}$. Which is likely to give me 12 critical points. Let's consider the $(0,0)$ first...
I expanded x around $x_0=y_0=0$ via $x=x_0+\alpha$. I reached that "$x'=x-x^3-xy^2$" linearizes as $\alpha ' \approx \alpha (1-3x_0^2-y_0^2)+x_0-x_0^3-x_0y_0^2-2x_0y_0\beta$. Here I see that I'm lucky that I chose first the critical point $(0,0)$ because the linearization reduces to $\alpha ' =\alpha$ and so $\alpha =Ae^t$. Right here I notice that when t grows up, so does alpha. So that I can already say that $(x,y)=(0,0)$ is not a stable critical point. Is my reasoning right?

P.S.:This is the very first exercise in the assignment of the course dealing with the study of critical points... Looks already quite complicated to me. :/

 

[fzero]

This looks ok so far.

 

[fluidistic]

This looks ok so far.

Ok! Very good to know. Thanks for having taught me how to tackle these kind of problems.