Stability Systems - Duffing oscillator In the case a=1,b=-1 this is the system:
     $$ dx=y $$
$$ dy=-x + x^3$$
I have to draw the phase space with the trajectories of the orbits. And I don´t know who to demonstrate the direction in the orbits. I only know is a circle for the $(0,0)$ and hyperbola for$(-1,0),(1,0)$.
 A: For the critical point at $(0, 0)$ just evaluate the dynamical system close to the origin. For example, take $x = 0.1$, and $y = 0$, you see that at that location ${\rm d y}/{\rm d}t < 0$, that means that a that location $y$ will be decreasing. In other words, the orbit going through $(0.1, 0)$ will rotate clock-wise. Same argument can be applied to the other critical points.
Here's a sketch to confirm it

A: Form the Jacobian of the system :
$$J(x,y) =  \begin{bmatrix} 0 & 1 \\ -1 + 3x^2 & 0\end{bmatrix}$$
For the origin $O(0,0)$ which is a critical point for the given system, it is :
$$J(0,0) = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$$
Then, the eigenvalues of the given Jacobian for the origin :
$$\det(J(0,0) -\lambda I) = 0 \Rightarrow \begin{vmatrix}  - \lambda & 1 \\ -1 & -\lambda\end{vmatrix} = 0 \Leftrightarrow \lambda^2 + 1 = 0 \Leftrightarrow \lambda = \pm i$$
This truly ensures that the origin $O(0,0)$ is a center for the given system and clockwise.
Now, you also have the critical points $A(-1,0)$ and $B(1,0)$. For $A$, the Jacobian is 
$$J(-1,0) = \begin{bmatrix} 0 & 1 \\ 2 & 0  \end{bmatrix}$$
with eigenvalues :
$$\det(J(-1,0) -\lambda I) = 0 \Rightarrow \begin{vmatrix}  - \lambda & 1 \\ 2 & -\lambda\end{vmatrix} = 0 \Leftrightarrow \lambda^2 -2 = 0 \Leftrightarrow \lambda = \pm \sqrt{2}$$
Since $\lambda_1 \cdot \lambda _2 < 0$ and the eigenvalues are purely real, the critical point $A$ will then be a saddle for the system, which by theory is unstable, thus the arrows are pointing away.
I will leave the case of $B$ up to you to work around.
A: Multiply
$$
\ddot x+x-x^3=0
$$
with $2\dot x$ and integrate to get
$$
\dot x^2+x^2-\frac12x^4=R^2
$$
and parametrize this as a circle equation to get polar coordinates
$$(R\cos \phi(t),R\sin(\phi))=\left(\dot x,x\sqrt{1-\frac12x^2}\right).$$
Close to $x=0$ the equation $u=f(x)=x\sqrt{1-\frac12x^2}$ in the second component is monotone and can be inverted as $$x=g(u)=g(R\sin(\phi))$$ with time derivative
$$
\dot x = g'(R\sin(\phi))R\cos \phi(t)\,\dot\phi(t)\implies 0=\dot x=R\cos \phi(t)\text{ or }1=g'(R\sin(\phi))\dot\phi(t),
$$
so that $$\dot\phi(t)=g'(R\sin(\phi))^{-1}=1+O(R^2).$$ This means counter-clock wise rotation in a $(\dot x,x)$ diagram or, diagonally reflected, clockwise rotation in a $(x,\dot x)$ diagram.
