Answer
$x^2+\dfrac{2}{3}x+\dfrac{1}{9}$
Work Step by Step
Using $(a+b)^2=a^2+2ab+b^2$ or the square of a binomial, the product of the given expression, $
\left( x-\dfrac{1}{3} \right)^2
,$ is
\begin{array}{l}\require{cancel}
(x)^2+2(x)\left( \dfrac{1}{3} \right)+\left( \dfrac{1}{3} \right)^2
\\\\=
x^2+\dfrac{2}{3}x+\dfrac{1}{9}
.\end{array}
