jmc

Writing the equation ${\log }_{64^{\frac{1}{3}}}{4}^x=1$ in exponential form gives $(64^{\frac{1}{3}}{)}^1=4^x$. This can be rewritten as $4^{3^{(\frac{1}{3})}}=4^x$, which means that $4^1=4^x$. Therefore, $x=\boxed{1}$.