Solve the following equation :

$4^{2x} = (\sqrt[3]{16})^{-\frac{6}{y}} = (\sqrt{8})^2$

Given,

$\Rightarrow 4^{2x} = (\sqrt[3]{16})^{-\frac{6}{y}} = (\sqrt{8})^2 \\[1em] \therefore 4^{2x} = (\sqrt{8})^2 \\[1em] \Rightarrow (2^2)^{2x} = 8 \\[1em] \Rightarrow (2)^{4x} = (2^3) \\[1em] \Rightarrow 4x = 3 \\[1em] \Rightarrow x = \dfrac{3}{4}.$

Similarly,

$\Rightarrow (\sqrt[3]{16})^{-\frac{6}{y}} = (\sqrt{8})^2 \\[1em] \Rightarrow (16)^{\frac{1}{3} \times -\frac{6}{y}} = 8 \\[1em] \Rightarrow (16)^{-\frac{2}{y}} = 8 \\[1em] \Rightarrow (2^4)^{-\frac{2}{y}} = 2^3 \\[1em] \Rightarrow (2)^{-\frac{8}{y}} = 2^3 \\[1em] \Rightarrow -\dfrac{8}{y} = 3 \\[1em] \Rightarrow y = -\dfrac{8}{3}.$

Hence, x = $\dfrac{3}{4} \text{ and } y = -\dfrac{8}{3}$.