(a) To find \(\frac{dy}{dx}\), we first find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

\(\frac{dx}{dt} = 1 + \frac{1}{t+2}\).

Using the product rule, \(\frac{dy}{dt} = e^{-2t} - 2(t-1)e^{-2t} = e^{-2t} - 2te^{-2t} + 2e^{-2t} = (3 - 2t)e^{-2t}\).

Now, \(\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{(3 - 2t)e^{-2t}}{1 + \frac{1}{t+2}} = \frac{(3 - 2t)(t + 2)}{t + 3} e^{-2t}\).

(b) A stationary point occurs when \(\frac{dy}{dx} = 0\).

Setting \((3 - 2t)(t + 2) = 0\), we solve for \(t\):

\(3 - 2t = 0 \Rightarrow t = \frac{3}{2}\).

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

\(y = \left(\frac{3}{2} - 1\right)e^{-3} = \frac{1}{2}e^{-3}\).

(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 \(3y = 2x - 1\).