Answer
$4y^2-2y+\dfrac{1}{4}$
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, $\left(
2y-\dfrac{1}{2}
\right)^2,$ is
\begin{array}{l}\require{cancel}
(2y)^2-2(2y)\left( \dfrac{1}{2} \right)+\left( \dfrac{1}{2} \right)^2
\\\\=
4y^2-2y+\dfrac{1}{4}
.\end{array}
