Given $x = a \sin 2t$, differentiate with respect to $t$:

$\frac{dx}{dt} = a \cdot \frac{d}{dt}(\sin 2t) = a \cdot (2\cos 2t) = 2a \cos 2t$

Given $y = a(\cos 2t + \log \tan t)$, differentiate with respect to $t$:

$\frac{dy}{dt} = a \cdot \frac{d}{dt}(\cos 2t + \log \tan t) = a \left( -2\sin 2t + \frac{1}{\tan t} \cdot \sec^2 t \right)$

Simplify the second term:

$\frac{1}{\tan t} \cdot \sec^2 t = \frac{\cos t}{\sin t} \cdot \frac{1}{\cos^2 t} = \frac{1}{\sin t \cos t} = \frac{2}{2\sin t \cos t} = \frac{2}{\sin 2t}$

So, $\frac{dy}{dt} = a \left( -2\sin 2t + \frac{2}{\sin 2t} \right) = 2a \left( \frac{1 - \sin^2 2t}{\sin 2t} \right) = 2a \left( \frac{\cos^2 2t}{\sin 2t} \right)$

$\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{2a \left( \frac{\cos^2 2t}{\sin 2t} \right)}{2a \cos 2t} = \frac{\cos^2 2t}{\sin 2t} \cdot \frac{1}{\cos 2t} = \frac{\cos 2t}{\sin 2t} = \cot 2t$