To evaluate the expression \(\left(\frac{81}{16}\right)^{\frac{-1}{4}}\times 2^{-1}\), we can start by simplifying the first term. We can rewrite \(\left(\frac{81}{16}\right)^{\frac{-1}{4}}\) as \(\left(\frac{3^4}{2^4}\right)^{\frac{-1}{4}}\), which can be further simplified to \(\left(\frac{3}{2}\right)^{-1}\). This is equal to \(\frac{2}{3}\).

Now, we can multiply this by the second term, \(2^{-1}\), which is equal to \(\frac{1}{2}\). So, the entire expression becomes \(\left(\frac{81}{16}\right)^{\frac{-1}{4}}\times 2^{-1} = \left(\frac{2}{3}\right) \times \left(\frac{1}{2}\right) = \frac{2}{6} = \frac{1}{3}\).