Answer
$\dfrac{1}{3}$ is not a zero of the function by the factor theorem.
Work Step by Step
The factor theorem states that when $f(a)=0$, then we have $(x-a)$ as a factor of $f(x)$ and when $(x-a)$ is a factor of $f(x)$, then $f(a)=0$.
We are given the function $f(x)=4x^3-5x^2-3x+1$
We simplify the given equation as follows:
$f(\dfrac{1}{3})=4(\dfrac{1}{3})^3-5(\dfrac{1}{3})^2-3(\dfrac{1}{3})+1 \\=\dfrac{-11}{27}$
This implies that $f(\dfrac{1}{3}) \ne 0$
So, $\dfrac{1}{3}$ is not a zero of the function by the factor theorem.
