Concept:
Use the hyperbolic identities \(\cosh^2 x - \sinh^2 x = 1\) and \(\tanh x = \frac{\sinh x}{\cosh x}\). Find \(\cosh^2 x\) and \(\sinh^2 x\) directly from \(\tanh x\), then compute the required expression.
Step 1: Express \(\cosh^2 x\) and \(\sinh^2 x\) in terms of \(\tanh x\).
Given \(\tanh x = \frac{1}{3}\). Let \(t = \tanh x\). Then \(\sinh^2 x = \frac{t^2}{1-t^2}\) and \(\cosh^2 x = \frac{1}{1-t^2}\).
\[
1 - t^2 = 1 - \frac{1}{9} = \frac{8}{9}.
\]
So,
\[
\cosh^2 x = \frac{1}{8/9} = \frac{9}{8}, \quad \sinh^2 x = \frac{1/9}{8/9} = \frac{1}{8}.
\]
Step 2: Compute \(\sinh^4 x\) and \(\cosh^4 x\).
\[
\sinh^4 x = \left(\frac{1}{8}\right)^2 = \frac{1}{64}, \quad \cosh^4 x = \left(\frac{9}{8}\right)^2 = \frac{81}{64}.
\]
Step 3: Evaluate \(E = 6\sinh^4 x + 2\cosh^4 x + 2\sinh^2 x + \cosh^2 x\).
\[
E = 6\left(\frac{1}{64}\right) + 2\left(\frac{81}{64}\right) + 2\left(\frac{1}{8}\right) + \frac{9}{8}.
\]
\[
= \frac{6}{64} + \frac{162}{64} + \frac{16}{64} + \frac{72}{64} = \frac{256}{64} = 4.
\]
Step 4: Write the final answer.
\[
\boxed{4}
\]