If $f(x)= x^{1/3} \sin x$. Say we want to find the derivative using the rules of differentiation.
So $f'(x)=\sin(x) x^{-2/3} + \cos(x) x^{1/3}$, and we see that $f'(0)$ is undefined.
But if we use the deriavtive defenition we seem to get zero. How can these be different?
The product rule states that for two differentiable functions $f,g$, we have that $\frac{d}{dx}fg(x) = f'(x)g(x) + f(x)g'(x)$. But in your example, $x^{1/3}$ is not differentiable at $0$, and so you cannot use the product rule to determine differentiability of $x^{1/3} \sin x$ at $x = 0$.
As you note in your post, using the definition of the derivative directly shows that $x^{1/3} \sin x$ is differentiable at $0$, and the derivative has value $0$ at $0$.