\[\left(9a^2-\frac{1}{16b^2}\right) = \left(3a - \frac{1}{4b}\right)\left(3a + \frac{1}{4b}\right)\]
\[\frac{\left(3a - \frac{1}{4b}\right)\left(3a + \frac{1}{4b}\right)}{\left(\frac{3a}{4b}-\frac{1}{4b}\right)} = \frac{\left(3a - \frac{1}{4b}\right)\left(3a + \frac{1}{4b}\right)}{\frac{1}{4b}(3a - 1)} \]
\[\frac{\left(3a - \frac{1}{4b}\right)\left(3a + \frac{1}{4b}\right)}{\frac{3a-1}{4b}} = \frac{\left(3a + \frac{1}{4b}\right)}{\frac{1}{4b}} = 4b\left(3a + \frac{1}{4b}\right) = 12ab + 1\]
\[12 \cdot \frac{2}{3} \cdot \left(-\frac{1}{12}\right) + 1 = -\frac{24}{36} + 1 = -\frac{2}{3} + 1 = \frac{1}{3}\]
Ответ: \(\frac{1}{3}\)