I’m having trouble understanding Lyapunov stability, which is
$\forall \epsilon > 0, \exists \delta > 0: \|\bar x(0)\| < \delta\implies \forall t \ge 0, \|\bar x(t)\| < \epsilon$
Why is the need for the $\delta$? Couldn’t we just say that both our equilibrium point and the state trajectory must remain in the $\epsilon$-ball?
That is, why shouldn't the definition be
$\forall \epsilon > 0, \|\bar x(0)\| < \epsilon \implies \forall t \ge 0, \|\bar x(t)\| < \epsilon$
Note that in the definition of Lyapunov stability, there are cases that $\|x(0)\| < \epsilon$ does not imply that $\|x(t)\| <\epsilon$. For example, consider the linear system
\begin{align} x' &= -2y, \\ y' &= x. \end{align}
Since $(x^2 + 2y^2)' = 2xx' + 4yy'= 0$, the solutions lies inside the ellipse
$$ x^2 + 2y^2 = c.$$
In particular, given any $\epsilon >0$, the solution with $\|x(0)\|<\epsilon$ might leave the ball with radius $\epsilon$. This system is Lyapunov stable, but not stable in the definition you proposed.
FYI: the stability that you proposed is equivalent to $x\cdot f(x) \le 0$ for all $x$ in a neighborhood of $0$.