Chapter 6 - Section 6.7 - Variation and Problem Solving - Exercise Set - Page 398: 31

$k=\dfrac{1}{3}$

Since $y$ varies directly as the cube of $x$, then, $ y=kx^3 .$ Substituting the given values of the variables, then the value of $k$ is, \begin{array}{l} 9=k(3)^3 \\\\ 9=k(27) \\\\ \dfrac{9}{27}=k \\\\ \dfrac{1}{3}=k \\\\ k=\dfrac{1}{3} .\end{array}