Consider the autonomous differential equation $$\frac{dx}{dt}=f(x(t)), \ \ \ x(t_0)=x_0 $$ where $f(x(t))=(x(t))^2(1-(x(t))^2)$, and let $x(t))$ be the displacement of a particle at time $t$.
The fixed points are $-1$, $0$ and $1$. Also $f'(-1)>0 $ so $x=-1$ is an unstable point. Conversely $f'(1)<0 $ so $x=1$ is a stable fixed point.
But my question is $f'(0)=0 $ so is $x=0$ stable or unstable? From drawing the direction field it looks as though if you start near $x=0$ you just move to $x=1$ but apparently if you start near $x=0$ then $x(t)\rightarrow 0 $. Why is this?
In 1D you may depict the stability by using arrows indicating the sign of the vector field outside of fixed points, whence in what direction a point moves with time. In your case $$ -\infty \longleftarrow (-1) \longrightarrow (0) \longrightarrow (1) \longleftarrow \infty $$ As Did mentions 0 appears stable from the left but unstable on the right. So the fixed point is neither stable nor unstable. This phenomena happens often when the derivative $\lambda = f'(x_0)$ of your vector field vanishes at a fixed point $f(x_0)=0$. Similar things happens in higher dimensions but with a larger variety of possible phenomena.