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Given $x = e^{\cos 3t}$. Differentiating with respect to $t$, we get: $$ \frac{dx}{dt} = e^{\cos 3t} \cdot \frac{d}{dt}(\cos 3t) = e^{\cos 3t} \cdot (-\sin 3t) \cdot 3 = -3e^{\cos 3t} \sin 3t $$ Since $x = e^{\cos 3t}$, we can write: $$ \frac{dx}{dt} = -3x \sin 3t $$
Given $y = e^{\sin 3t}$. Differentiating with respect to $t$, we get: $$ \frac{dy}{dt} = e^{\sin 3t} \cdot \frac{d}{dt}(\sin 3t) = e^{\sin 3t} \cdot (\cos 3t) \cdot 3 = 3e^{\sin 3t} \cos 3t $$ Since $y = e^{\sin 3t}$, we can write: $$ \frac{dy}{dt} = 3y \cos 3t $$
Using the chain rule, we have: $$ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{3y \cos 3t}{-3x \sin 3t} = -\frac{y \cos 3t}{x \sin 3t} $$
We have $x = e^{\cos 3t}$, so $\ln x = \cos 3t$. Similarly, $y = e^{\sin 3t}$, so $\ln y = \sin 3t$. Substituting these into the expression for $\frac{dy}{dx}$, we get: $$ \frac{dy}{dx} = -\frac{y \ln x}{x \ln y} $$
Final Answer: $\frac{dy}{dx} = -\frac{y \ln x}{x \ln y}$