I am trying to proof that the following equation is stable if $|1-ln(\alpha)|<1$;
$$f(x)=\alpha x e^{- \beta x}$$
So the equilibrity points would be $x=0$ and $x=\frac{ln\alpha}{\beta}$, the derivate of the equation would be:
$$f '(x)= \alpha e^{-\beta x} (1-\beta \alpha)$$
Now i don't know how to proceed neither how to explain why the equilibrium would be stable.
So you are looking for local stability of the mapping $x_{k+1}=f(x_k)$.
As you pointed there are two equilibrium points $x=0$ and $x=\ln(\alpha)/\beta$. The mapping $f$ can be written as
$x_{k+1}=f(x_{eq})+f'(x_{eq})(x_k-x_{eq})+o(|x_k-x_{eq}|)=x_{eq}+f'(x_{eq})(x_k-x_{eq})+o(|x_k-x_{eq}|)$.
Equivalently we have
$x_{k+1}-x_{eq}=f'(x_{eq})(x_k-x_{eq})+o(|x_k-x_{eq}|)$.
Thus the equilibrium point $x_{eq}$ is stable (locally) if $|f'(x_{eq})|<1$ (first theorem of Lyapunov -discrete case) that yields your desired condition $|f'(\ln(\alpha)/\beta)|=|1-\ln\alpha|<1$ for the equilibrium $x_{eq}=\ln(\alpha)/\beta$.
Edit: The condition $|f'(x_{eq})|<1$ can be derived if we consider the metric $V_k=(x_k-x_{eq})^2$. Then, $V_{k+1}-V_k=- [1-(f'(x_{eq}))^2](x_k-x_{eq})^2+o(|x_k-x_{eq}|^2)\leq 0$ if $ |f'(x_{eq})|<1$ and $|x_k-x_{eq}|$ small enough.