Answer
$\left(\dfrac{1}{4}-y\right)\left(\dfrac{1}{4}+y\right)$
Work Step by Step
Using $a^2-b^2=(a+b)(a-b)$ or the factoring of the difference of two squares, then,
\begin{align*}
\dfrac{1}{16}-y^2
\Rightarrow
\left(\dfrac{1}{4}-y\right)\left(\dfrac{1}{4}+y\right)
.\end{align*}