#### Answer

$-\frac{11}{b^{2}}$

#### Work Step by Step

Using the rule $\frac{x^{a}}{x^{b}}=x^{a-b}$, we obtain:
$\frac{-44a^{3}b^{4}}{4a^{3}b^{6}}$
$\frac{-44}{4}\times\frac{a^{3}}{a^{3}}\times\frac{b^{4}}{b^{6}}$
$=-11\times a^{3-3}\times b^{4-6}$
$=-11\times a^{0}\times b^{-2}$
$=-11\times 1\times \frac{1}{b^{2}}$
$=\frac{-11}{b^{2}}$
$=-\frac{11}{b^{2}}$