The following scalar system:
$$\dot{x} = x^{2} - x^{3} $$
has two equilibrium points, one located at $x = 0$ and the other at $x = 1$. After looking at the phase portrait, setting xdot(2,1) = 0 and xdot(1,1) = x(1)^2 - x(1)^3) it is possible to say that both equilibrium points are asymptotic stable.
Knowing this, how can I obtain a valid Lyapunov function that can guarantee asymptotic stability, since just using:
\begin{align} V(x) &= \frac{1}{2}x^{2}\\ \dot{V} &= x^{3} - x^{4} \end{align}
Can only guarantee that $V$ is positive definite, but cannot be used to guarantee that $\dot{V}$ is negative definite.
You have a scalar system, and there is no phase portrait for it, it lives on $\mathbb{R}$. And two equilibrium points connected by a feasible trajectory cannot be asymptotically stable simultaneously.
For your system, trajectories starting at $x_0<0$ converge to zero, and trajectories starting with $x_0>0$ converge to 1. The equilibrium point $x=1$ is locally asymptotically stable.
More precisely, let $z=x-1$ and consider $V = \frac{1}{2}z^2$. Then $\dot{z} = (z+1)^2-(z+1)^3 = -z(z+1)^2$ and $$\dot{V} = -z^2(z+1)^2.$$ You observe that $\dot V<0$ for all $z$ except for $\dot{V}=0$ at $z=-1$ and $z=0$. So trajectories always move towards $z=0$ (i.e., $x=1$), but they stuck at $z=-1$ (i.e., $x=0$).