Setting up the difference quotient at x = 0

Plugging into the definition, $f'(0) = \lim_{h\to 0} \dfrac{h^{2/3} - 0}{h} = \lim_{h\to 0} \dfrac{h^{2/3}}{h}$. Since $h^{2/3} = (h^{1/3})^2 \geq 0$, we can simplify this as $\lim_{h\to 0} h^{2/3 - 1} = \lim_{h\to 0} h^{-1/3} = \lim_{h\to 0} \dfrac{1}{h^{1/3}}$.

As $h \to 0^+$, $h^{1/3} \to 0^+$, so $\dfrac{1}{h^{1/3}} \to +\infty$. As $h \to 0^-$, $h^{1/3} \to 0^-$, so $\dfrac{1}{h^{1/3}} \to -\infty$. Since the two one-sided limits disagree (and neither is finite), the limit does not exist, so $f'(0)$ does not exist and $f(x) = x^{2/3}$ is not differentiable at $x=0$ — this reflects the vertical tangent (cusp) in its graph there.

The tangent line behavior approaches vertical near zero. This is because the difference quotient $\dfrac{1}{h^{1/3}}$ grows without bound as $h$ approaches zero from either side, going to positive infinity from the right and negative infinity from the left. This unbounded steepness is exactly why the limit defining the derivative fails to exist there. It reflects the sharp vertical cusp visible in the graph at that point.