What is $\left(\frac{6}{7}\right)^2 \cdot \left(\frac{1}{2}\right)^2$?
Since $\left(\frac{a}{b}\right)^j \cdot \left(\frac{c}{d}\right)^j = \left(\frac{a \cdot c}{b \cdot d}\right)^{j}$, we know  $\left(\frac{6}{7}\right)^2 \cdot \left(\frac{1}{2}\right)^2 = \left(\frac{6 \cdot 1}{7 \cdot 2}\right)^2$.  Simplifying, we have $\left(\frac{3}{7}\right)^2 = \frac{3^2}{7^2}$, because $\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}$.  We know $3^2 = 9$ and $7^2 = 49$, so our answer is $\boxed{\frac{9}{49}}$.