$\frac{d}{dx}(\sin(x^{\frac{1}{3}}))$ from first principle

You have

$$\begin{aligned} \frac{\sin(x^{\frac{1}{3}}) - \sin(c^{\frac{1}{3}}) }{x-c} &= \frac{2\sin\left(\frac{x^{\frac{1}{3}}- c^{\frac{1}{3}}}{2}\right)\cos\left(\frac{x^{\frac{1}{3}}+c^{\frac{1}{3}}}{2}\right)}{x-c}\\ &=\frac{2\sin\left(\frac{x^{\frac{1}{3}}- c^{\frac{1}{3}}}{2}\right)\cos\left(\frac{x^{\frac{1}{3}}+c^{\frac{1}{3}}}{2}\right)}{(x^{\frac{1}{3}} - c^{\frac{1}{3}})(x^{\frac{2}{3}} + c^{\frac{2}{3}} + x^{\frac{1}{3}}c^{\frac{1}{3}})} \end{aligned}$$

Now as $$\lim\limits_{x \to c }\frac{2\sin\left(\frac{x^{\frac{1}{3}}- c^{\frac{1}{3}}}{2}\right)}{x^{\frac{1}{3}} - c^{\frac{1}{3}}} = 1$$

we get $$\frac{d}{dx}(\sin(x^{\frac{1}{3}})) = \frac{\cos(x^{\frac{1}{3}})}{3 x^{\frac{2}{3}}}$$

hint

Replace in

$$\frac{a-b}{a^3-b^3}=\frac{1}{a^2+ab+b^2}$$

$ a \;\; \text{by} \;\; (x+h)^\frac 13 $

and

$ b\;\; \text{by}\;\; x^\frac 13$.