Answer
$\pm 1;\pm 2;\pm 3;\pm 6;\pm 9;\pm 18;\pm \dfrac {1}{3};\pm \dfrac {2}{3}$
Work Step by Step
In a polinomial function like
$f\left( x\right) =a_{n}x^{n}+a_{n-1}x^{n-1}+\ldots a_{1}x+a_{0}$
If $p/q$, in lowest terms, is a rational zero of $f$, then p must be a factor of $a_0 $ and $q$ must be a factor of $a_n$.
Here
$f\left( x\right) =3x^{5}-x^{2}+2x+18\Rightarrow a_{n}=3;a_{0}=18 $
Factors of $a_0$ are $\pm 1,\pm 2,\pm 3,\pm6, \pm 9,\pm18$
Factors of $a_n$ are $\pm 1,\pm 3$
So the potential rational zeros are
$\pm 1;\pm 2;\pm 3;\pm 6;\pm 9;\pm 18;\pm \dfrac {1}{3};\pm \dfrac {2}{3}
$
