Local center manifold Consider the system
$$
x'=x^2, \quad  y'=y
$$
and find the local center manifold of the equilibrium $(0,0)$.
The linearization matrix is
$$
A=\begin{pmatrix}0 & 0\\0 & 1\end{pmatrix},
$$
hence the eigenvalues are $\lambda_1=0$ and $\lambda_2=1$.
The center space is spanned by the eigenvector belonging to $\lambda_1$, say $v=(1,0)$. Hence the $x$-axis should be the center space.
But what is the center manifold?
Since the center manifold $W_c$ should be tangent to the x-axis in point (0,0), my idea would be to write
$$
W_c=\left\{(x,y): y=h(x)\right\}, \text{with }h(0)=0, h'(0)=0
$$
and then to make some ansatz like second order approximation:
$$
h(x)=ax²+O(\lvert x\rvert³).
$$
Do we then get that $h(x)=0$ is a second order approximation of $W_c$, i.e. $W_c$ is the x-axis?
 A: For your system, the center manifold coincides with the $x-$axis, i.e. $$W^c = \{(x,y) \in  \mathbb{R}^2 \, | \, y = 0\}$$ Thus basically $h(x) \equiv 0$. Indeed, let $x=x(t)$ be a solution of the equation $\dot{x} = x^2$. Then you can check directly that $(x(t), 0 )$ is a solution of 
\begin{align}
\dot{x} &= x^2\\
\dot{y} &= y
\end{align}
When expanding in terms of series, you are looking for an analytic approximation of your center manifold. However, your manifold $W^c$ is already analytic, so you will keep matching it. The rest of the center manifolds are not analytic at the origin, so you will keep missing them. The other center manifolds are continuous extensions of the graphs $y = Ce^{-\frac{1}{x}}$ (if I have computed them correctly) so not analytic at $x=0$.  They are infinitely smooth, bit their Taylor coefficients at $0$ are all $0$. In any case, you will keep getting zero as coefficients.
Edit. The other center manifolds of the system  \begin{align}
\dot{x} &= x^2\\
\dot{y} &= y
 \end{align}
are in fact orbits of the system, i.e. they are curves traced by the solutions of near the origin. These curves can be written in the form $y = Ce^{-\frac{1}{x}}$ (one can calulate them explicitly). So if you take a solutions $(x(t),y(t))$ of
 \begin{align}
\dot{x} &= x^2\\
\dot{y} &= y
\end{align} near the origin, there exists a real constant $C$ depending on the initial conditions, such that $y(t) = Ce^{-\frac{1}{x(t)}}$ for all $t \in (-\infty, \epsilon)$. Thus the one dimensional manifold $y = Ce^{-\frac{1}{x}}$ is invariant manifold. However, it is not defined for $x=0$. Nevertheless, we can extend it by letting  $y = h_C(x) =  Ce^{-\frac{1}{x}}$ for $x \neq 0$ and $y=h_C(0) = 0$ for $x=0$. Now $h_C(x)$ defines a center manifold $W^c_C$, $h_C$ is also continuous at $x=0$ and one can see by direct calculation that all of its derivatives are also continuous at $x=0$ and are equal to $0$. Hence  $W^c_C$ is a infinitely smooth manifold and it's function $h_C(x)$ is infinitely differentiable at $x=0$. Thus we can conclude that all of its Taylor coefficients at $0$ are equal to zero. $W^c_C$ is not an a real analytic submanifold of the plane. However the special center manifold $W^c = \{y=0\}$ is the only real analytic center manifold. And that is why you are approximating $W^c$ and not the rest of the center manifolds. In order to derive the equation $y = C e^{-\frac{1}{x}}$ just writ down the system as
\begin{align}
\frac{dx}{dt} &= x^2\\
\frac{dy}{dt} &= y
\end{align} and then rewrite it in terms of differentials as
\begin{align}
\frac{dx}{x^2} &= dt\\
\frac{dy}{y} &= dt
\end{align}
Now equate the first and the second equation, because $dt =dt$, and get
\begin{align}
 \frac{dy}{y} = dt = \frac{dx}{x^2} 
\end{align} that is
\begin{align}
 \frac{dy}{y} = \frac{dx}{x^2}  
\end{align}
which is exact differential and 
$$d\big( \log(y)\big) = d \left(- \frac{1}{x}\right) $$ which after integration of both parts turns into
$$\log(y)  = D - \frac{1}{x} $$ and when you exponentiate both sides of the equation
$$y = e^D \, e^{-\frac{1}{x}}$$ Now let $C=d^D$ and allow $C$ to be any real number. You get $$y = C e^{-\frac{1}{x}}$$ Now direct verification, shows that for any solution $( x(t), y(t) )$
$$\frac{dy}{dt}(t)  = C e^{-\frac{1}{x(t)}} \left(\frac{1}{x(t)^2}\right) \frac{dx}{dt}(t)$$ for all $t$.    
