\[\begin{cases} \left(\frac{1}{8}\right)^{3x} \cdot \left(\frac{1}{8}\right)^{-y} = \frac{1}{64} \\ \log_{2} 8x - \log_{2} y = 2 \end{cases}\]

\[\begin{aligned} x &= \square;\\ y &= \square. \end{aligned}\]