Start with the given equation:

\(\ln(y + 1) - \ln y = 1 + 3 \ln x\)

Use the logarithm quotient rule: \(\ln \left( \frac{y+1}{y} \right) = 1 + 3 \ln x\)

Exponentiate both sides to remove the logarithm:

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

Use the property \(e^{a+b} = e^a \cdot e^b\):

\(\frac{y+1}{y} = e \cdot (e^{\ln x})^3\)

Since \(e^{\ln x} = x\), this becomes:

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

Rearrange to solve for \(y\):

\(y + 1 = e x^3 y\)

\(1 = e x^3 y - y\)

\(1 = y(e x^3 - 1)\)

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

Thus, \(y = (e^{x^3} - 1)^{-1}\).