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

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

Substitute $x = \pi$ into the given equation:
$y^{\cos \pi} = \pi^{\sin 2\pi}$
$y^{-1} = \pi^0$
$y^{-1} = 1 \implies y = 1$

2. Take the natural logarithm of both sides to handle variable exponents.

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

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

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

4. Substitute $x = \pi$ and $y = 1$ to find the exact value of $\frac{\mathrm{d}y}{\mathrm{d}x}$.

$-\sin(\pi) \ln(1) + \cos(\pi) \left(\frac{1}{1}\right)\frac{\mathrm{d}y}{\mathrm{d}x} = 2\cos(2\pi) \ln \pi + \sin(2\pi) \left(\frac{1}{\pi}\right)$
$-0 \cdot 0 + (-1)\frac{\mathrm{d}y}{\mathrm{d}x} = 2(1) \ln \pi + 0 \cdot \frac{1}{\pi}$
$-\frac{\mathrm{d}y}{\mathrm{d}x} = 2\ln \pi$
$\frac{\mathrm{d}y}{\mathrm{d}x} = -2\ln \pi$