Separate the variables:

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

Integrate both sides:

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

The left-hand side becomes:

\(\frac{1}{3} y^3\)

For the right-hand side, use integration by parts:

Let \(u = x\) and \(dv = 6e^{3x} \, dx\).

Then \(du = dx\) and \(v = 2e^{3x}\).

\(\int 6x e^{3x} \, dx = 2xe^{3x} - \int 2e^{3x} \, dx\)

\(= 2xe^{3x} - \frac{2}{3}e^{3x} + C\)

Equating both sides:

\(\frac{1}{3} y^3 = 2xe^{3x} - \frac{2}{3}e^{3x} + C\)

Use the initial condition \(y = 2\) when \(x = 0\):

\(\frac{1}{3} (2)^3 = 2(0)e^{0} - \frac{2}{3}e^{0} + C\)

\(\frac{8}{3} = 0 - \frac{2}{3} + C\)

\(C = \frac{10}{3}\)

Thus, the equation becomes:

\(\frac{1}{3} y^3 = 2xe^{3x} - \frac{2}{3}e^{3x} + \frac{10}{3}\)

Substitute \(x = 0.5\) to find \(y\):

\(\frac{1}{3} y^3 = 2(0.5)e^{1.5} - \frac{2}{3}e^{1.5} + \frac{10}{3}\)

Calculate \(y\) to 2 decimal places:

\(y \approx 2.44\)