Answer
$R=2$
$x+\dfrac {1}{3}$ is not a factor.
Work Step by Step
$f\left( x\right) =3x^{4}+x^{3}-3x+1=\left( x+\dfrac {1}{3}\right) q\left( x\right) +R$
$\Rightarrow f\left( -\dfrac {1}{3}\right) =2=R\neq 0$
Since $R\ne0$, $x+\dfrac {1}{3}$ is not a factor.
