Given
\( \rm (3^x + \frac{1}{3^x} = 9 \frac{1}{9})\)

Let \( \rm  a = 3^x\). Substituting the value of \(a\) in the equation gives us:

\(\rm a + \frac{1}{a} = 9 \frac{1}{9}\)

\( \rm a + \frac{1}{a} = \frac{9 \cdot 9 + 1}{9}\)

\( \rm a + \frac{1}{a} = \frac{82}{9}\)

Multiplying both sides of the equation by \(a\), we get:

\( \rm a \cdot a + \frac{1}{a} \cdot a = \frac{82}{9} \cdot a\)

\( \rm a^2 + 1 = \frac{82a}{9}\)

Multiplying both sides of the equation by \(9\), we get:

\( \rm 9a^2 + 9 = 82a\)

Subtracting \(82a\) from both sides of the equation, we get:

\( \rm 9a^2 - 82a + 9 = 82a - 82a\)

\( \rm 9a^2 - 82a + 9 = 0\)

The above equation is quadratic in \(a\). We use the factorization method to solve it:

\(\rm 9a^2 - (81 + 1)a + 9 = 0\)

\(\rm 9a^2 - 81a - a + 9 = 0\)

\(\rm 9a (a - 9) - 1 (a - 9) = 0\)

\(\rm (9a - 1)(a - 9) = 0\)

Either:

\(\rm 9a - 1 = 0\)

\(\rm 9a = 1\)

\(\rm a = \frac{1}{9}\)

Substituting the value, \(a = 3^x\), we get:

\(\rm 3^x = \frac{1}{3^2}\)

\(\rm 3^x = 3^{-2}\)

\(\rm \therefore x = -2\)

Or:

\(\rm a - 9 = 0\)

\(\rm a = 9\)

Substituting the value, \(a = 3^x\), we get:

\(\rm 3^x = 3^2\)

\(\rm \therefore x = 2\)

Hence, the required values of \(x\) are {-2, 2}.