If ${\mathrm{xy}}^2=4\text{ and }{\log }_3\left({\log }_2\mathrm{x}\right)+{\log }_{1/3}\left({\log }_{1/2}\mathrm{y}\right)=1\text{, }$ then x=

$\begin{array}{l}{\log }_3\left({\log }_2\mathrm{x}\right)+{\log }_{1/3}\left({\log }_{1/2}\mathrm{y}\right)=1\\ \text{ or }{\log }_3\left({\log }_2\mathrm{x}\right)-{\log }_3\left({\log }_{1/2}\mathrm{y}\right)=1\\ \text{ or }{\log }_3\left({\log }_2\left(4/{\mathrm{y}}^2\right)\right)-{\log }_3\left({\log }_{1/2}\mathrm{y}\right)=1\\ \text{ or }{\log }_2\left(4/{\mathrm{y}}^2\right)=3\left({\log }_{1/2}\mathrm{y}\right)\\ \text{ or }{\log }_2\left(4/{\mathrm{y}}^2\right)=-3\left({\log }_2\mathrm{y}\right)\\ \text{ or }{\log }_2\left(4/{\mathrm{y}}^2\right)+\left({\log }_2{\mathrm{y}}^3\right)=0\\ \text{ or }4\mathrm{y}=1\\ \text{ or }\mathrm{y}=1/4\\ \Rightarrow \mathrm{x}=64\end{array}$