#### Answer

$\dfrac{5}{4y}$

#### Work Step by Step

RECALL:
$x^{-m}=\dfrac{1}{x^m},m \gt 0$
Use the rule above to have:
$\\=\dfrac{1}{y^1}+\dfrac{1}{(4y)^1}
\\=\dfrac{1}{y}+\dfrac{1}{4y}$
Make the expressions similar using the LCD which is $4y$ to have:
$\\=\dfrac{1(4)}{y(4)} + \dfrac{1}{4y}
\\=\dfrac{4}{4y} + \dfrac{1}{4y}
\\=\dfrac{5}{4y}$