From the second given equation: $\log_2 y = 3 \implies y = 2^3 = 8$.

Using the first equation: $\log_y x = 4 \implies x = y^4 = (2^3)^4 = 2^{12}$.

Now, evaluate the argument $\frac{x}{y^2}$: $\frac{x}{y^2} = \frac{2^{12}}{(2^3)^2} = \frac{2^{12}}{2^6} = 2^6$.

We need to find $\log_4\left(\frac{x}{y^2}\right)$: $\log_4 (2^6) = \log_{2^2} (2^6) = \frac{6}{2} \log_2 2 = 3$.

Hence, the correct option is $3$.