Answer
No, it's not a zero. See work for explantion
Work Step by Step
$f(\frac{1}{3})=2(\frac{1}{3})^3+3(\frac{1}{3})^2-6(\frac{1}{3})+7=\frac{2}{27}+\frac{3}{9}-\frac{6}{3}+7=5.417$
$5.417\ne0$
Therefore $\frac{1}{3}$ is not a zero of the function.
