Answer
$\left(\dfrac{1}{4}x^2+y^2\right)\left(\dfrac{1}{2}x+y\right)\left(\dfrac{1}{2}x-y\right)$
Work Step by Step
Using the factoring of a difference of $2$ squares which is given by $a^2-b^2=(a+b)(a-b),$ the given $\text{
expression,
}$ $
\dfrac{1}{16}x^4-y^4
,$ is equivalent to
\begin{align*}
&
\left(\dfrac{1}{4}x^2\right)^2-(y^2)^2
\\\\&=
\left(\dfrac{1}{4}x^2+y^2\right)\left(\dfrac{1}{4}x^2-y^2\right)
.\end{align*}
Since the second factor above is still a difference of $2$ squares, then it can be further factored using the same method above. That is,
\begin{align*}
&
\left(\dfrac{1}{4}x^2+y^2\right)\left[\left(\dfrac{1}{2}x\right)^2-(y)^2\right]
\\\\&=
\left(\dfrac{1}{4}x^2+y^2\right)\left[\left(\dfrac{1}{2}x+y\right)\left(\dfrac{1}{2}x-y\right)\right]
\\\\&=
\left(\dfrac{1}{4}x^2+y^2\right)\left(\dfrac{1}{2}x+y\right)\left(\dfrac{1}{2}x-y\right)
.\end{align*}
Hence, the given expression simplifies to $
\left(\dfrac{1}{4}x^2+y^2\right)\left(\dfrac{1}{2}x+y\right)\left(\dfrac{1}{2}x-y\right)
$.
