Answer
Domain: $(-\infty, 1/3)\cup(1/3, \infty)$
Range: $(-\infty, 0)\cup(0, \infty)$
Work Step by Step
Let $f(x) = \frac{2}{3x-1}$
This function is defined for all values of $x$, except when $3x-1 = 0 \implies 3x = 1 \implies x = \frac{1}{3}$
Since this value cancels the denominator, therefore the function is not defined for $x = \frac{1}{3}$
Thus, the domain of $f = (-\infty, 1/3)\cup(1/3, \infty)$
Given $f(x) = y$
$y = \frac{2}{3x-1} \implies y(3x-1) = 2 \implies 3xy-y = 2 \implies 3xy = y + 2
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\implies x = \frac{y+2}{3y}$
This expression is defined for all values of $y$, except when $3y = 0 \implies y = 0$, for the same reasoning as above.
Therefore, the range of $f = (-\infty, 0)\cup(0, \infty)$
See the attached figure.
