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

\(x = (\ln t)^2\) implies \(\frac{dx}{dt} = 2 \ln t \cdot \frac{1}{t} = \frac{2 \ln t}{t}\).

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

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

Substitute \(t = e\):

\(\frac{dy}{dx} = -\frac{e^3 e^{2-e^2}}{2 \ln e} = -\frac{e^3 e^{2-e^2}}{2}\).

Simplifying gives \(-e^{4-e^2}\).