To find the gradient of the curve at the point where it crosses the y-axis, we first need to determine the value of \(t\) when \(x = 0\).

Since \(x = \\ln(2t + 3)\), setting \(x = 0\) gives:

\(\\ln(2t + 3) = 0\)

\(2t + 3 = 1\)

\(2t = -2\)

\(t = -1\)

Now, we find \(\frac{dy}{dx}\) using the chain rule:

\(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}\)

First, find \(\frac{dx}{dt}\):

\(x = \\ln(2t + 3)\)

\(\frac{dx}{dt} = \frac{2}{2t + 3}\)

Next, find \(\frac{dy}{dt}\):

\(y = \frac{3t + 2}{2t + 3}\)

Using the quotient rule:

\(\frac{dy}{dt} = \frac{(2t + 3)(3) - (3t + 2)(2)}{(2t + 3)^2}\)

\(= \frac{6t + 9 - (6t + 4)}{(2t + 3)^2}\)

\(= \frac{5}{(2t + 3)^2}\)

Now, substitute \(t = -1\) into \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{\frac{5}{(2(-1) + 3)^2}}{\frac{2}{2(-1) + 3}}\)

\(= \frac{\frac{5}{1^2}}{\frac{2}{1}}\)

\(= \frac{5}{2}\)

Thus, the gradient of the curve at the point where it crosses the y-axis is \(\frac{5}{2}\).