College Algebra, Chapter 3, 3.1, Section 3.1, Problem 64

Determine the domain of the function $\displaystyle f(x) = \frac{x}{\sqrt[4]{9-x^2}}$
The function is not defined when the radicand is a negative value and when the denominator is 0. So,

$\begin{equation} \begin{aligned} 9 - x^2 &> 0 && \text{Difference of squares}\\ \\ (3+x)(3-x) &> 0 \end{aligned} \end{equation}$


The factors on the left hand side of the inequality are $3+x$ and $3-x$. These factors are zero when $x$ is $-3$ and $3$, respectively. These numbers divide the number line into intervals
$(-\infty,-3),(-3,3),(3,\infty)$

By testing some points on the interval



Thus, the domain of $f(x)$ is $(-3,3)$