I'm not sure how to solve this: $\frac{5}{9}x^\frac{2}{3}$. I applied the product rule and have $\frac{2}{3}\frac{5}{9}x^{-\frac{1}{3}}$.
$\frac{30}{9}x^{-\frac{1}{3}}$, then $\frac{9}{30}x^{\frac{1}{3}}$.
This isn't the answer in the book, though. Is the procedure for getting rid of a negative in the exponent to flip the number in front of $x$?
The comments have already pointed out where you have misunderstood this question.
However, as an answer to your question:
\begin{align*}\frac{\mathrm{d}}{\mathrm{d}x}\Bigg(\frac{5}{9}x^\frac{2}{3} \Bigg)&= \frac{5}{9}\frac{\mathrm{d}}{\mathrm{d}x}(x^\frac{2}{3})\\ &= \frac{5}{9}\Bigg(\frac{2}{3}\Bigg)x^{\frac{2}{3}-1}\\ &= \frac{10}{27}x^{-\frac{1}{3}} \end{align*}