Answer
$\dfrac{\sqrt[3]{y}}{y^3}$
Work Step by Step
Recall the rational exponent property (pg. 382):
$a^{\frac{m}{n}}=\sqrt[n]{a^m}=(\sqrt[n]{a})^m$
Applying this property, we get:
$\dfrac{\sqrt[9]{y^3}}{\sqrt[3]{y^9}}=\dfrac{y^{\frac{3}{9}}}{y^{\frac{9}{3}}}=\dfrac{y^{\frac{1}{3}}}{y^{3}}$
Recall the rational exponent property (pg. 383):
$\dfrac{a^m}{a^n}=a^{m-n}$
Applying this property to the expression above, we get:
\begin{align*}
\dfrac{y^{\frac{1}{3}}}{y^{3}}&=y^{\frac{1}{3}-3}\\
&=y^{\frac{1}{3}-\frac{9}{3}}\\
&=y^{\frac{1-9}{3}}\\
&=y^{\frac{-8}{3}}\\
\end{align*}
Next, recall the basic property of negative exponents (pg. 383):
$a^{-m}=\frac{1}{a^m}$
Using the rule above gives:
$y^{\frac{-8}{3}}=\dfrac{1}{y^{\frac{8}{3}}}$
Now, we rewrite the result in radical form:
$\dfrac{1}{y^{\frac{8}{3}}}=\dfrac{1}{\sqrt[3]{y^8}}$
Finally, we get rid of the radical in the denominator by multiplying $\sqrt[3]{y}$ to both the numerator and the denominator:
\begin{align*}
\dfrac{1}{\sqrt[3]{y^8}}&=\dfrac{1 \cdot \sqrt[3]{y}}{\sqrt[3]{y^8} \cdot \sqrt[3]{y}}\\\\
&=\dfrac{\sqrt[3]{y}}{\sqrt[3]{y^9}}\\\\
&=\dfrac{\sqrt[3]{y}}{y^3}
\end{align*}
