$\begin{align}  & {{\left( \frac{81}{16} \right)}^{-\tfrac{1}{4}}}\times {{2}^{-1}}={{\left( \frac{{{3}^{4}}}{{{2}^{4}}} \right)}^{-\tfrac{1}{4}}}\times {{2}^{-1}}={{\left( \frac{3}{2} \right)}^{4\times (-\tfrac{1}{4})}}\times {{2}^{-1}} \\ & \text{     }={{\left( \frac{3}{2} \right)}^{-1}}\times {{2}^{-1}}=\frac{2}{3}\times \frac{1}{2}=\frac{1}{3} \\\end{align}$