How can I know if an equilibrium is stable in a difference equation? Suppose that I have the following equation:
$$
y_{t+1} = \frac{2+y_t-y^3_t}{2y_t}
$$
Then
$$
\bar{y} = \frac{2+\bar{y} -\bar{y}^3}{2\bar{y}}
$$
And so I obtain these solutions: $$y_1 = -2, \quad y_2 = -1, \quad y_3 = 1 $$
Now how can I know if these equilibria are stable?
I believe that it is useful to compute the derivative with respect to the solutions found.
Therefore I obtain the derivative:
$$
\frac{d}{dy}
=
-\frac{(y^3+1)}{y^2}
$$
For $y_1 = -2$ the derivative becomes: $\frac{7}{4}$
For $y_2 = -1$ the derivative becomes: $0$
For $y_3 = 1$ the derivative becomes: $-2$
So how can I know if the equilibria found are stable?
 A: What you have is not a differential equation, but rather a difference equation. The criterion for stability of the fixed points is then slightly different:
Let $x_{n+1}=F(x_n)$ be a map for which we find fixed points by solving $F(x^{*})=x^{*}$.There are three cases you need to check (as with differential equations):

*

*If $|F'(x^{*})|<1$,the fixed point is stable

*If $|F'(x^{*})|>1$,the fixed point is unstable

*If $|F'(x^{*})|=1$,we cannot decide whether it is stable or unstable.This is called a "border" case.

For difference equations in 2D/3D, you can analogously check for stability, but the derivative is replaced by a Jacobian matrix, where you linearize and check whether the complex modulus of the eigenvalues is less than 1.
Edit: To answer your question, then the only fixed point which is stable is $y_2$. The other ones are unstable.
Edit: Perhaps I should provide a justification of why this is true. You can prove it just like the case when you have a differential equation.
Let $\Delta x_{n}=x_n-x^{*}$ be a small displacement of the fixed point. Taylor expand the map about $x^{*}$ to get:
$$x^*+\Delta x_{n+1}=F(x^*+\Delta x_n)=F(x^*)+F'(x^*)\Delta x_n+... $$
Because $F(x^*)=x^*$, then you can actually simplify the above to:
$$\Delta x_{n+1}=F'(x^*)\Delta x_n+... $$
The idea is then that for arbitrarily small $\Delta x_n$, you can negate the higher order terms. You can then see that the 3 cases I listed above are what we need to check for stability.
Edit: What I mean by 2D, for instance, is a system of the form:
$$\begin{cases} x_{n+1}=F(x_n,y_n) \\ y_{n+1}=G(x_n,y_n) \end{cases}$$
To study stability in this case, you set $F(x^*,y^*)=x^*$ and $G(x^*,y^*)=y^*$ to find the fixed points.
You then find the Jacobian matrix evaluated at that point:
$$J(x^*,y^*) = \begin{bmatrix}
    \frac{\partial F}{\partial x} &  \frac{\partial F}{\partial y} \\
    \frac{\partial G}{\partial x} &  \frac{\partial G}{\partial y}
\end{bmatrix}\Bigg\rvert_{(x^*,y^*)}$$
To study stability, you find the eigenvalues of this matrix and then you check whether the complex modulus is less than 1. That is, both roots must lie inside the unit circle in order for the fixed point to be stable. Note that you have to repeat this process for different fixed points.
You can do the same for higher dimensional systems. However, you typically don't go higher than 3D for physical applications.
