As Victor says:

I am confident that the system
$$ x' = -y + ( x^2 + y^2) x, $$
$$ y' = x + ( x^2 + y^2) y $$
has the origin repelling nearby trajectories, while
$$ x' = -y - ( x^2 + y^2) x, $$
$$ y' = x - ( x^2 + y^2) y $$
has the origin attracting nearby trajectories, and
$$ x' = -y $$
$$ y' = x $$
has just periodic orbits near the origin. But all three
linearize to the same thing at the origin,
$$
\left( \begin{array}{rr}
0 & -1 \\\
1 & 0
\end{array}
\right) .
$$
with eigenvalues $\pm i.$

EDIT: Indeed, given a constant real number $\lambda$ and system
$$ x' = -y + \lambda ( x^2 + y^2) x, $$
$$ y' = x + \lambda ( x^2 + y^2) y , $$
we find that
$$ \frac{d}{dt} \; (x^2 + y^2) = 4 \lambda (x^2 + y^2)^2. $$

EDIT some more: so, for the nonconstant paths, if we set time to $0$ when the trajectory crosses the unit circle, we get
$$ x^2 + y^2 = \frac{1}{1 - 4 \lambda t} $$
showing that when $\lambda > 0$ the path reaches infinite radius in finite time, while with
$\lambda < 0$ the path spirals in to the origin, as expected.
Then, if we set $$ x = r \cos \theta, \; y = r \sin \theta $$
as usual, the rate of change of $ \theta $ does not depend on $ \lambda $ and $ \forall \lambda,t$ we have
$$ \frac{d \theta}{d t} = 1. $$