Answer: \(x=\sqrt3\), \(y=3^{1/3}\).

We start with the main method. Use logarithm laws and index laws first so that the equation is reduced to an algebraic equation.

Let

\(u=\log_3x\quad\text{and}\quad v=\log_3y.\)

Use change of base within powers of \(3\):

\(\log_{81}y=\frac{\log_3y}{\log_381}=\frac{v}{4}\)

and

\(\log_9x=\frac{\log_3x}{\log_39}=\frac{u}{2}.\)

The first equation becomes

\(8u+12\left(\frac v4\right)=5.\)

So

\(8u+3v=5.\)

The second equation becomes

\(4\left(\frac u2\right)+3v=2.\)

So

\(2u+3v=2.\)

Subtract the second equation from the first:

\(6u=3.\)

Therefore

\(u=\frac12.\)

Substitute into \(2u+3v=2\):

\(2\left(\frac12\right)+3v=2.\)

So

\(1+3v=2\), and hence

\(v=\frac13.\)

Now convert back:

\(\log_3x=\frac12\), so \(x=3^{1/2}=\sqrt3.\)

Also

\(\log_3y=\frac13\), so \(y=3^{1/3}.\)

This completes the solution and gives the required result.