Answer
$\displaystyle \frac{6y^{3}}{x^{7}}$
Work Step by Step
Multiplication is commutative (we can change the order of factors.)
It is also associative (we can choose what to multiply first.)
$(\displaystyle \frac{1}{3}x^{-5}y^{4})(18x^{-2}y^{-1})$=$(\displaystyle \frac{1}{3}\cdot 18)(x^{-5}\cdot x^{-2})(y^{4}\cdot y^{-1})$
... use the rule:$\quad a^{m}\cdot a^{n}=a^{m+n}$
$=6x^{-5-2}y^{4-1}$
$=6x^{-7}y^{3}\quad $... use the rule: $a^{-n}=\displaystyle \frac{1}{a^{n}}$
= $\displaystyle \frac{6y^{3}}{x^{7}}$
