To find the derivative of $f(x) = \sqrt[3]{x}$ using the definition of the derivative, we apply the limit definition:

$f'(x) = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$

Here, $f(x) = \sqrt[3]{x}$, so:

[ f'(x) = \lim_{\Delta x \to 0} \frac{\sqrt[3]{x + \Delta x} - \sqrt[3]{x}}{\Delta x} ]

To simplify this expression, we use the identity for the difference of cubes:

$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$

Setting $a = \sqrt[3]{x + \Delta x}$ and $b = \sqrt[3]{x}$, the expression becomes:

[ \left(\sqrt[3]{x + \Delta x} - \sqrt[3]{x}\right)\left(\left(\sqrt[3]{x + \Delta x}\right)^2 + \sqrt[3]{x + \Delta x} \cdot \sqrt[3]{x} + \left(\sqrt[3]{x}\right)^2\right) = x + \Delta x - x = \Delta x ]

So, we have:

[ \Delta x = \left(\sqrt[3]{x + \Delta x} - \sqrt[3]{x}\right)\left(\left(\sqrt[3]{x + \Delta x}\right)^2 + \sqrt[3]{x + \Delta x} \cdot \sqrt[3]{x} + \left(\sqrt[3]{x}\right)^2\right) ]

This gives us:

[ \frac{\Delta x}{\sqrt[3]{x + \Delta x} - \sqrt[3]{x}} = \left(\left(\sqrt[3]{x + \Delta x}\right)^2 + \sqrt[3]{x + \Delta x} \cdot \sqrt[3]{x} + \left(\sqrt[3]{x}\right)^2\right) ]

Now, substituting into the original limit:

[ f'(x) = \lim_{\Delta x \to 0} \frac{1}{\left(\sqrt[3]{x + \Delta x}\right)^2 + \sqrt[3]{x + \Delta x} \cdot \sqrt[3]{x} + \left(\sqrt[3]{x}\right)^2} ]

As $\Delta x$ approaches 0, $\sqrt[3]{x + \Delta x}$ approaches $\sqrt[3]{x}$, so:

[ f'(x) = \frac{1}{3