Answer
The domain is $ \{(-\infty,\infty) \cap (-\infty , 3) \}= (-\infty , 3) $ .
Work Step by Step
$$
f(x)=\sqrt{\frac{x^{2}}{3-x}}
$$
the values $x$ for $f(x)$ is defined when:
$$
x^{2} \geq 0 \quad \text {and} \quad (3-x) \gt 0
$$
$ \Rightarrow $
$$
-\infty \lt x \lt \infty \quad \text {and} \quad x \lt 3
$$
since the radical cannot be negative and the denominator of the function
cannot be zero.
So we can observe that only values in the intervals $(-\infty, 3) $ satisfy the inequality.
So the domain is $(-\infty,\infty) \cap (-\infty , 3) = (-\infty , 3) $ .
