First, separate the variables:

\(\frac{dy}{y} = \frac{1}{(x+1)(3x+1)} dx\)

Integrate both sides:

\(\int \frac{dy}{y} = \int \frac{1}{(x+1)(3x+1)} dx\)

\(\ln |y| = \int \left(\frac{A}{x+1} + \frac{B}{3x+1}\right) dx\)

Using partial fraction decomposition, find \(A\) and \(B\):

\(A = -\frac{1}{2}, \quad B = \frac{3}{2}\)

Integrate:

\(\ln |y| = -\frac{1}{2} \ln |x+1| + \frac{1}{2} \ln |3x+1| + C\)

Use the initial condition \(y = 1\) when \(x = 1\) to find \(C\):

\(\ln 1 = -\frac{1}{2} \ln 2 + \frac{1}{2} \ln 4 + C\)

\(C = \frac{1}{2} \ln 2\)

Substitute \(x = 3\) into the equation:

\(\ln |y| = -\frac{1}{2} \ln 4 + \frac{1}{2} \ln 10 + \frac{1}{2} \ln 2\)

\(\ln |y| = \frac{1}{2} \ln 5\)

\(|y| = e^{\frac{1}{2} \ln 5} = \sqrt{5}\)

Thus, \(y = \frac{1}{2} \sqrt{5}\).