Answer: \(x\lt -\dfrac12\) or \(x\gt 2\).

Consider the two cases for the modulus.

If \(3x-1\geq0\), then \(|3x-1|=3x-1\). The inequality becomes

\(3x-1\gt 3+x.\)

So

\(2x\gt 4,\)

and hence

\(x\gt 2.\)

If \(3x-1\lt 0\), then \(|3x-1|=1-3x\). The inequality becomes

\(1-3x\gt 3+x.\)

So

\(-4x\gt 2,\)

giving

\(x\lt -\dfrac12.\)

Therefore

\(\boxed{x\lt -\dfrac12\ \text{or}\ x\gt 2}.\)