Answer
(i) $\dfrac{f(x)}{g(x)}=4x^3-3x^2+1-\frac{1}{3x}$
(ii) The domain of $\dfrac{f(x)}{g(x)}$ does not include 0.
Work Step by Step
Divide $f(x)$ by $g(x)$ to have:
$\dfrac{f(x)}{g(x)} = \dfrac{12x^4-9x^3+3x-1}{3x}$
$\dfrac{f(x)}{g(x)} = 4x^3-3x^2+1-\frac{1}{3x}$
Since $3x$ is in the denominator of the original quotient, and a denominator is not allowed to be zero, then
$3x \ne 0
\\x \ne 0$
