Answer
$R=0$. $x+3$ is a factor
Work Step by Step
$f\left( x\right) =2x^{6}-18x^{4}+x^{2}-9=\left( x+3\right) q\left( x\right) +R$
$\Rightarrow f\left( -3\right) =2\times \left( -3\right) ^{6}-18\times \left( -3\right) ^{4}+\left( -3\right) ^{2}-9=0=R$
Since $R=0$, $x+3$ is a factor.