Answer
$x^2+\dfrac{1}{3}x+\dfrac{1}{36}
\Rightarrow\left(
x+\dfrac{1}{6} \right)^2
$
Work Step by Step
The third term of a perfect square trinomial is equal to the square of half the coefficient of the middle term. Hence, to complete the square of the given expression $
x^2+\dfrac{1}{3}x
,$ the third term must be
\begin{array}{l}\require{cancel}\left(
\dfrac{1/3}{2}\right)^2
\\\\=
\left(
\dfrac{1}{6}\right)^2
\\\\=
\dfrac{1}{36}
.\end{array}
Using $a^2\pm2ab+b^2=(a\pm b)^2$, then
\begin{array}{l}\require{cancel}
x^2+\dfrac{1}{3}x+\dfrac{1}{36}
\Rightarrow\left(
x+\dfrac{1}{6} \right)^2
.\end{array}
