Answer
The domain is $(-\infty,-1) \cup (\frac{1}{3},\infty)$ .
Work Step by Step
$$
f(x)=\frac{1}{\sqrt {3x^{2}+2x-1}}=\frac{1}{\sqrt {(3x-1)(x+1)}}
$$
the values $x$ for $f(x)$ is defined when:
$$
(3x-1)(x+1) \gt 0
$$
since the radical cannot be negative and the denominator of the function
cannot be zero.
So we can solve:
$$
(3x-1)(x+1) = 0
$$
$ \Rightarrow $
$$
x_{1}=\frac{1}{3},\:x_{2}=-1
$$
this numbers $-1$ and $ \frac{1}{3}$ divide the number line
into 3 intervals, $ (-\infty,-1), (-1 ,\frac{1}{3}) $ and $(\frac{1}{3},\infty)$.
we observe that only values in the intervals $(-\infty,-1) $ and $(\frac{1}{3},\infty)$ satisfy the inequality.
So the domain is $(-\infty,-1) \cup (\frac{1}{3},\infty)$ .
