Let \(y = 3^x\). Then \(3^{-x} = \frac{1}{y}\).

The equation becomes \(\frac{y^2 + \frac{1}{y}}{y^2 - \frac{1}{y}} = 4\).

Multiply numerator and denominator by \(y\) to clear the fraction:

\(\frac{y^3 + 1}{y^3 - 1} = 4\).

Cross-multiply to get:

\(y^3 + 1 = 4(y^3 - 1)\).

Expand and simplify:

\(y^3 + 1 = 4y^3 - 4\).

Rearrange terms:

\(3y^3 = 5\).

So, \(y^3 = \frac{5}{3}\).

Since \(y = 3^x\), we have \((3^x)^3 = \frac{5}{3}\).

Thus, \(3^{3x} = \frac{5}{3}\).

Take the logarithm of both sides:

\(3x \log 3 = \log \left( \frac{5}{3} \right)\).

Solve for \(x\):

\(x = \frac{\log \left( \frac{5}{3} \right)}{3 \log 3}\).

Calculate \(x\) to 3 decimal places:

\(x \approx 0.155\).