(a) Differentiate \(x\) and \(y\) with respect to \(t\):

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

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

The gradient \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2}{3(2+3t)}\).

Since \(2+3t > 0\), the gradient is always positive.

(b) At the y-axis, \(x = 0\), so \(\ln(2+3t) = 0\) implies \(2+3t = 1\), hence \(t = -\frac{1}{3}\).

Substitute \(t = -\frac{1}{3}\) into \(y = \frac{t}{2+3t}\):

\(y = \frac{-\frac{1}{3}}{1} = -\frac{1}{3}\).

The point is \((0, -\frac{1}{3})\).

The gradient at this point is \(\frac{2}{3}\).

The equation of the tangent is \(y + \frac{1}{3} = \frac{2}{3}x\).

Rearrange to get \(3y = 2x - 1\).