Separate variables and factorize to obtain:

\(\frac{dy}{(3y+1)(y+3)} = 4x \, dx\)

Express the left side as partial fractions:

\(\frac{A}{3y+1} + \frac{B}{y+3}\)

Solving for \(A\) and \(B\), we find \(A = \frac{3}{8}\) and \(B = -\frac{1}{8}\).

Integrate both sides:

\(\int \left( \frac{3}{8(3y+1)} - \frac{1}{8(y+3)} \right) dy = \int 4x \, dx\)

This gives:

\(\frac{1}{8} \ln(3y+1) - \frac{1}{8} \ln(y+3) = 2x^2 + c\)

Substitute \(x = 0\) and \(y = 1\) to find \(c\):

\(\frac{1}{8} \ln(4) - \frac{1}{8} \ln(4) = c\)

\(c = 0\)

Rearrange to solve for \(y\):

\(\ln \left( \frac{3y+1}{y+3} \right) = 16x^2\)

\(\frac{3y+1}{y+3} = e^{16x^2}\)

\(3y + 1 = (y + 3)e^{16x^2}\)

\(3y + 1 = ye^{16x^2} + 3e^{16x^2}\)

\(y(3 - e^{16x^2}) = 3e^{16x^2} - 1\)

\(y = \frac{3e^{16x^2} - 1}{3 - e^{16x^2}}\)