The function given is:

$g(x) = \frac{\sqrt[3]{x}}{x^2 + 9}$

To determine the domain of this function, we need to consider the following:

1. Numerator Analysis: The numerator is $\sqrt[3]{x}$. The cube root function is defined for all real numbers, so there are no restrictions from the numerator.

2. Denominator Analysis: The denominator is $x^2 + 9$. Since this is a quadratic function $x^2$ plus a positive constant (9), it is always positive for all real numbers. Therefore, $x^2 + 9$ never equals zero, and there is no restriction on $x$ from the denominator.

Since both the numerator and the denominator are defined for all real numbers, the function $g(x)$ is defined for all $x \in \mathbb{R}$.

Conclusion: The domain of $g(x)$ is all real numbers. In interval notation, this is:

$\text{Domain of } g(x): \ (-\infty, \infty)$

Would you like further details on this problem, or do you have any questions?

Here are five related questions to consider:

1. What is the range of the function $g(x)$?
2. How would the domain change if the numerator were $\sqrt{x}$ instead of $\sqrt[3]{x}$?
3. What happens to the domain if the denominator includes a variable term in the numerator (e.g., $x^2 - 9$)?
4. How do you find the domain of a function that includes a square root in the denominator?
5. What are the asymptotes, if any, of the function $g(x)$?

Tip: When analyzing the domain of a function, always consider where the function could become undefined, such as division by zero or even roots of negative numbers.