Answer
$16y^2-2y+\dfrac{1}{16}$
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, $\left(
4y-\dfrac{1}{4}
\right)^2,$ is
\begin{array}{l}\require{cancel}
(4y)^2-2(4y)\left( \dfrac{1}{4} \right)+\left( \dfrac{1}{4} \right)^2
\\\\=
16y^2-2y+\dfrac{1}{16}
.\end{array}
