Differentiability at x = -2 via the limit definition

Since $f(-2) = 0$, the difference quotient becomes $\dfrac{f(-2+h)-f(-2)}{h} = \dfrac{h^{4/3}}{h} = h^{1/3}$. This uses the fact that $f(-2+h) = ((-2+h)+2)^{4/3} = h^{4/3}$, and simplifying the exponents gives $h^{4/3-1} = h^{1/3}$.

Now examine the one-sided limits. For the right-hand limit: $\lim_{h\to 0^+} h^{1/3} = 0$. For the left-hand limit, using the real cube root function defined for all real numbers: $\lim_{h\to 0^-} h^{1/3} = 0$. Since both one-sided limits exist and are equal, the two-sided limit exists: $\lim_{h\to 0} h^{1/3} = 0$. Therefore $f'(-2) = 0$ exists, meaning $f$ is differentiable at $x=-2$. Graphically, this corresponds to a horizontal tangent line at that point (not a corner, cusp, or vertical tangent).

The tangent line is horizontal. In part a, the derivative was computed as the limit of h to the one third power, which equals zero. Since the derivative exists and equals zero, the tangent line has slope zero, meaning it is horizontal rather than vertical.