Clearly, domain of $f ( x )$ is $R$ and $f ( x ) >0, \forall x \in R$
Let $y=f(x)$
Then, $y=\frac{1}{2-\cos 3 x}$
$\Rightarrow \cos 3 x=\frac{2 y-1}{y}$
$\Rightarrow x=\frac{1}{3} \cos ^{-1}\left(\frac{2 y-1}{y}\right)$
For $x$ to be real, we must have
$-1 \leq \frac{2 y-1}{y} \leq 1$
$\Rightarrow -y \leq 2 y-1 \leq y$
$\Rightarrow -y \leq 2 y-1 \text { and } 2 y-1 \leq y$
$\Rightarrow 3 y \geq 1 \text { and } y \leq 1$
$\Rightarrow y \geq \frac{1}{3} \text { and } y \leq 1$
$\Rightarrow y \in\left[\frac{1}{3}, 1\right]$
Therefore, range $(f)=\left[\frac{1}{3}, 1\right]$