Answer
$\dfrac{1}{\sqrt[8]{y^9}}$ or $\dfrac{1}{(\sqrt[8]{y})^9}$
Work Step by Step
Recall the basic property of negative exponents (pg. 383):
$a^{-m}=\frac{1}{a^m}$
Applying this property, we get:
$y^{-\frac{9}{8}}=\dfrac{1}{y^{\frac{9}{8}}}$
Next, recall the rational exponent property (pg. 382):
$a^{\frac{m}{n}}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m$
Applying this property ($m=9$, $n=8$, $a=y$), we get:
$\dfrac{1}{y^{\frac{9}{8}}}=\dfrac{1}{\sqrt[8]{y^9}}$ or $\dfrac{1}{(\sqrt[8]{y})^9}$
