Since $f(x) = x^{1/3}$ has an infinite derivative at $x=0$, we don't expect to have bounded difference quotients on any interval containing $x=0.$ In fact, we can be led to this expectation by using only precalculus ideas. Since the graph of $y = x^3$ straightens out horizontally at the origin, the graph of the inverse function (of $y = x^3$), namely that graph of $y = x^{1/3},$ straightens out vertically at the origin.
Thus, we're led to consider difference quotients based at the origin:
$\frac{f(x) - f(0)}{x-0} \; = \; \frac{x^{1/3}}{x} \; = \; \frac{1}{x^{2/3}}$
Since these difference quotients are not bounded on any interval containing $x=0$, for example the interval $[0,1],$ the function $f(x)$ is not Lipschitz continuous on any interval containing the origin.
More generally, $f(x)$ is not Lipschitz on any $E \subseteq {\mathbb R}$ such that $0$ is a limit point of $E.$ For this last result, simply note that the values of
$\frac{f(r) - f(s)}{r-s} \; = \; \frac{1}{r^{2/3} + (rs)^{1/3} + s^{2/3}}$
will be unbounded if both $r$ and $s$ can be chosen arbitrarily close to $0.$