Given that $y^{\cos x} = x^{\sin 2x}$, find the exact value of $\frac{dy}{dx}$ at $x = \pi$.

1. Find the corresponding value of $y$ when $x = \pi$.

When $x = \pi$,
$y^{\cos \pi} = \pi^{\sin 2\pi}$
$y^{-1} = \pi^0$
$y^{-1} = 1$
$y = 1$

2. Apply the natural logarithm to both sides to simplify the exponents.

$\ln(y^{\cos x}) = \ln(x^{\sin 2x})$
$\cos x \ln y = \sin 2x \ln x$

3. Differentiate implicitly with respect to $x$ using the product rule on both sides.

$\frac{d}{dx}(\cos x \ln y) = \frac{d}{dx}(\sin 2x \ln x)$
$-\sin x \ln y + \cos x \left(\frac{1}{y}\right)\frac{dy}{dx} = 2\cos 2x \ln x + \sin 2x \left(\frac{1}{x}\right)$

4. Substitute $x = \pi$ and $y = 1$ to solve for the exact value of $\frac{dy}{dx}$.

$-\sin(\pi) \ln(1) + \cos(\pi) \left(\frac{1}{1}\right)\frac{dy}{dx} = 2\cos(2\pi) \ln \pi + \sin(2\pi) \left(\frac{1}{\pi}\right)$
$-(0)(0) + (-1)(1)\frac{dy}{dx} = 2(1)\ln \pi + (0)\left(\frac{1}{\pi}\right)$
$-\frac{dy}{dx} = 2\ln \pi$
$\frac{dy}{dx} = -2\ln \pi$