The given function is:

$y = \cos\left(e^{-3\theta^2}\right)$

We are asked to find the derivative of this function with respect to $\theta$.

Steps for differentiation:

1. Apply the chain rule: Since this is a composition of functions (a cosine function with an exponential function inside), we need to apply the chain rule.

2. Differentiate the outer function: The derivative of $\cos(u)$ with respect to $u$ is $-\sin(u)$.

   So: $\frac{dy}{d\theta} = -\sin\left(e^{-3\theta^2}\right) \cdot \frac{d}{d\theta}\left(e^{-3\theta^2}\right)$

3. Differentiate the inner function: Now, we differentiate $e^{-3\theta^2}$ with respect to $\theta$. Using the chain rule again: $\frac{d}{d\theta}\left(e^{-3\theta^2}\right) = e^{-3\theta^2} \cdot \frac{d}{d\theta}(-3\theta^2)$ The derivative of $-3\theta^2$ is $-6\theta$, so: $\frac{d}{d\theta}\left(e^{-3\theta^2}\right) = e^{-3\theta^2} \cdot (-6\theta)$

4. Combine the results: Substituting back into the original derivative, we get: $\frac{dy}{d\theta} = -\sin\left(e^{-3\theta^2}\right) \cdot e^{-3\theta^2} \cdot (-6\theta)$

   Simplifying the expression: $\frac{dy}{d\theta} = 6\theta \cdot \sin\left(e^{-3\theta^2}\right) \cdot e^{-3\theta^2}$

Thus, the derivative is: $\frac{dy}{d\theta} = 6\theta \sin\left(e^{-3\theta^2}\right) e^{-3\theta^2}$

Would you like further details or have any other questions?

Here are some related questions to further your understanding:

1. How do you apply the chain rule to more complex composite functions?
2. What is the derivative of $\sin(e^{x^2})$?
3. How does the chain rule apply to trigonometric functions?
4. How would you handle the derivative of a product of two functions?
5. Can you apply the chain rule to integrals?

Tip: Always differentiate the outer function first, then work your way inward with the chain rule when dealing with composite functions.