Answer
$\color{blue}{\dfrac{1}{2pq}}$
Work Step by Step
RECALL:
(1) $a^{-m} = \dfrac{1}{a^m}$
(2) $\dfrac{a^m}{a^n} = a^{m-n}$
(3) $\left(\dfrac{a}{b}\right)^m=\dfrac{a^m}{b^m}$
(4) $(ab)^m = a^mb^m$
(5) $(a^m)^n=a^{mn}$
(6) $a^m \cdot a^n = a^{m+n}$
(7) $a^0=1, a\ne0$
Use rule (6) above to obtain:
$=\dfrac{3pq^{1+2}}{6p^2q^4}
\\=\dfrac{3pq^3}{6p^2q^4}$
Divide the coefficients by canceling out the common factors to obtain:
$\require{cancel}
=\dfrac{\cancel{3}pq^3}{\cancel{6}2p^2q^4}
\\=\dfrac{pq^3}{2p^2q^4}$
Use rule (2) above to obtain:
$=\dfrac{1}{2} \cdot p^{1-2} \cdot q^{3-4}
\\=\dfrac{1}{2} \cdot p^{-1} \cdot q^{-1}$
Use rule (1) above to obtain:
$=\dfrac{1}{2} \cdot \dfrac{1}{p^1} \cdot \dfrac{1}{q^1}
\\=\dfrac{1}{2} \cdot \dfrac{1}{p} \cdot \dfrac{1}{q}
\\=\color{blue}{\dfrac{1}{2pq}}$
