Answer
$9x^2+2x+\dfrac{1}{9}$
Work Step by Step
Using $(a\pm b)^2=a^2\pm2ab+b^2$ or the square of a binomial, the square of the given expression, $\left(
3x+\dfrac{1}{3}
\right)^2,$ is
\begin{array}{l}\require{cancel}
(3x)^2+2(3x)\left( \dfrac{1}{3} \right)+\left( \dfrac{1}{3} \right)^2
\\\\=
9x^2+2x+\dfrac{1}{9}
.\end{array}
