To find the gradient \(\frac{dy}{dx}\), we first need to find \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\).

For \(x = \frac{t}{2t + 3}\), use the quotient rule:

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

For \(y = e^{-2t}\), differentiate with respect to \(t\):

\(\frac{dy}{dt} = -2e^{-2t}\).

Now, use the chain rule to find \(\frac{dy}{dx}\):

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

Evaluate at \(t = 0\):

\(\frac{dy}{dx} = -2e^{0} \cdot \frac{3^2}{3} = -2 \cdot 3 = -6\).