Let's solve this problem where $y = \sqrt[3]{x}$ and find $y'$.

Step 1: Express the function in exponential form

The cube root of $x$ can be written as: $y = \sqrt[3]{x} = x^{\frac{1}{3}}$

Step 2: Differentiate with respect to $x$

To find $y'$, differentiate both sides with respect to $x$: $\frac{dy}{dx} = \frac{d}{dx} \left( x^{\frac{1}{3}} \right)$

Step 3: Apply the power rule

The power rule for differentiation is $\frac{d}{dx} \left( x^n \right) = n \cdot x^{n-1}$. Applying this to our function: $y' = \frac{1}{3} \cdot x^{\frac{1}{3} - 1} = \frac{1}{3} \cdot x^{-\frac{2}{3}}$

Step 4: Simplify the expression

$y' = \frac{1}{3} \cdot \frac{1}{x^{\frac{2}{3}}} = \frac{1}{3} x^{-\frac{2}{3}}$

This corresponds to one of the choices provided.

The correct answer is: $\boxed{\frac{1}{3} x^{-\frac{2}{3}}}$

Would you like more details or have any questions? Here are some related questions:

1. How does the power rule apply to fractional exponents?
2. What is the derivative of $y = \sqrt[4]{x}$?
3. How can we find the derivative of a more complex root function like $y = \sqrt[5]{x^2}$?
4. How does negative exponent notation simplify understanding derivatives?
5. What happens if you differentiate the function $y = \frac{1}{x}$?
6. How do you differentiate products or quotients of functions?
7. Can you provide a physical interpretation of the derivative for this function?
8. What is the process for finding higher-order derivatives of $y = \sqrt[3]{x}$?

Tip: When dealing with roots, always express them as fractional exponents to simplify the differentiation process.