First, separate the variables:

\(\frac{dy}{y} = \frac{6e^{3x}}{2 + e^{3x}} \, dx\).

Integrate both sides:

\(\int \frac{1}{y} \, dy = \int \frac{6e^{3x}}{2 + e^{3x}} \, dx\).

The left side integrates to \(\ln y\).

For the right side, use substitution: let \(u = 2 + e^{3x}\), then \(du = 3e^{3x} \, dx\), so \(dx = \frac{du}{3e^{3x}}\).

Thus, \(\int \frac{6e^{3x}}{u} \cdot \frac{du}{3e^{3x}} = 2 \int \frac{1}{u} \, du = 2 \ln |u| + C\).

Substitute back \(u = 2 + e^{3x}\):

\(\ln y = 2 \ln |2 + e^{3x}| + C\).

Exponentiate both sides:

\(y = e^{C} (2 + e^{3x})^2\).

Given \(y = 36\) when \(x = 0\):

\(36 = e^{C} (2 + e^{0})^2 = e^{C} \cdot 3^2 = 9e^{C}\).

Thus, \(e^{C} = 4\).

Therefore, \(y = 4(2 + e^{3x})^2\).