Center manifold theorem example. Does somebody know an easy way to tackle these center manifold problems?

Consider the system
  \begin{align}
\dot x&=ax^3+x^2y\\
\dot y &=-y+y^2+xy-x^3
\end{align}
  (a) Determine an approximation for the Center Manifold of this system.
(b) Use Center Manifold theory to investigate the stability of the origin $(x,y)=(0,0)$ depending on the parameter $a$ (so for $a<0,a>0$ and $a=0$).

It has to be written in some standard form I guess:
\begin{align}
\dot{x}&=Cx + F(x,y)
\\
\dot{y}&=Py+G(x,y)
\end{align}
In this case this leads to:
$C=0$, $\quad$ $P=-1$, $\quad$ $F(x,y)=ax^{3}+x^{2}y$ $\quad$ and 
$G(x,y)=y^2+xy-x^3$. 
Next we define an $h(x)$ as:
$$h(x)=a_2x^2+a_3x^3+a_4x^4+...$$
We take $\frac{\partial}{\partial x}h(x) = 2a_2x+3a_3x^2+4a_4x^3+...$ and substitute this into:
$$\frac{\partial}{\partial x}h(x)(Cx+F(x,h(x)))=Ph(x)+G(x,h(x))$$ 
which gives:
$$(2a_2x+3a_3x^2+4a_4x^3+...)(ax^3+x^2h(x)) \\ =-(a_2x^2+a_3x^3+a_4x^4+...)+(h(x)^2+xh(x)-x^3)$$
Which leads to:
$$2aa_2x^4+3aa_3x^5+4aa_4x^6+2a_2h(x)x^3+3a_3h(x)x^4+4a_4h(x)x^5
\\
=-a_2h(x)^2x^2-a_2h(x)x^3+a_3h(x)^2x^3+a_3h(x)x^4+a_4h(x)^2x^4-a_2x^5+a_4h(x)x^5-a_3x^6-a_4x^7$$
And then take $O(x^2)$ and $O(x^3)$  and leave out the other terms.
I really doubt if this is the way to do it.
 A: For small $x$, the first equation tells us that $x$ moves slowly compared to the exponential decay of $y$. Now the (moving) equilibrium of the second equation is at the roots of that quadratic equation, giving roots at $y\approx 1-x$ and $y\approx -x^3$. The first is unstable and far away from $(0,0)$, the second tells us that the series for $h(x)$ starts with $-x^3$.
$$
h(x)=-x^3+c_4x^4+c_5x^5+...
$$
Inserted into the DE $h'(x)F(x,h(x))=-h(x)+G(x,h(x))$ results then in the lower power terms
\begin{align}
&x^5(-3+4c_4x+5c_5x^2+...)(a-x^2+c_4x^3+c_5x^4+....)
\\&=x^3-c_4x^4-c_5x^5-...+x^6(-1+c_4x+c_5x^2+...)^2+x^4(-1+c_4x+c_5x^2+...)-x^3
\end{align}
The third power terms on the right cancel. The 4th power terms give
$$
0=-c_4-1\implies c_4=-1.
$$
In the fifth power we get terms on both sides
$$
-3a=-c_5+c_4\implies c_5=3a-1
$$
Continuing to the 6th degree
$$
4c_4a=-c_6+1+c_5\implies c_6=7a
$$
and so on.

The slow flow along the center manifold follows approximately the differential equation
$$
\dot x = ax^3+x^2h(x)=ax^3-x^5-x^6+(3a-1)x^7+7ax^8+...
$$
as the fast flow in $y$ direction will quickly revert any deviations towards the curve $y=h(x)$.
For medium to large $a$ the stability is thus determined by the sign of $a$ while for small $a$ the picture can be refined to contain the next term 
$$
\dot x=x^3(a-x^2)
$$
which gives a pitchfork bifurcation in direction of positive $a$ with stable equilibria at about $x=\pm\sqrt{a}$ and an unstable point at the origin.
