Chapter 11 - Polynomial and Rational Functions - 11.1 Power Functions and Proportionality - Exercises and Problems for Section 11.1 - Exercises and Problems - Page 439: 3

Yes, $g(x)=-\frac{1}{6}x^9$

Use the fact that $(-x^3)^3=(-1 \cdot x^3)^3$, $(ab)^n=a^n(b^n)$, and $(a^m)^n=a^{mn}$ to get $$g(x)=\frac{(-1)^3(x^3)^3}{6}=-\frac{x^9}{6}=-\frac{1}{6}x^9$$ Therefore, this is a power function with $k=-\frac{1}{6}$ and $p=9$.