(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 \(y = (t - 1)e^{-2t}\):

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