Answer
$\pm9,\pm3,\pm\frac{3}{2},\pm 1;\pm\frac{1}{2},\pm \frac {1}{3},\pm\frac{9}{2},\pm\frac{1}{6}$
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) =6x^{4}-x^{2}+9\Rightarrow a_{n}=6;a_{0}=9 $
Factors of $a_0$ are $\pm 1,\pm3,\pm9$
Factors of $a_n$ are $\pm1 ,\pm2, \pm3,\pm6$
So the potential rational zeros are:
$\pm9,\pm3,\pm\frac{3}{2},\pm 1;\pm\frac{1}{2},\pm \frac {1}{3},\pm\frac{9}{2},\pm\frac{1}{6}$