We start with the equation: \[ \left(\frac{1}{9}\right)^{x} = 81 \]
We can express \(\frac{1}{9}\) as \(9^{-1}\) and \(81\) as \(9^2\): \[ (9^{-1})^x = 9^2 \]
Using the property of exponents \((a^m)^n = a^{mn}\), we get: \[ 9^{-x} = 9^2 \]
Since the bases are the same, we can set the exponents equal to each other: \[ -x = 2 \]
Solving for \(x\), we get: \[ x = -2 \]
\[ \boxed{x = -2} \]
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