First, find \(\frac{dx}{d\theta}\) and \(\frac{dy}{d\theta}\):

\(x = \tan \theta \Rightarrow \frac{dx}{d\theta} = \sec^2 \theta\).

\(y = 2 \cos^2 \theta \sin \theta\).

Using the product rule, \(\frac{dy}{d\theta} = 2 \left(2 \cos \theta (-\sin \theta) \sin \theta + \cos^2 \theta \cos \theta\right) = -4 \cos \theta \sin^2 \theta + 2 \cos^3 \theta\).

Now, find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}} = \frac{-4 \cos \theta \sin^2 \theta + 2 \cos^3 \theta}{\sec^2 \theta}\).

\(\frac{dy}{dx} = (-4 \cos \theta \sin^2 \theta + 2 \cos^3 \theta) \cos^2 \theta\).

Using \(\sin^2 \theta = 1 - \cos^2 \theta\), substitute:

\(-4 \cos \theta (1 - \cos^2 \theta) + 2 \cos^3 \theta = -4 \cos \theta + 4 \cos^3 \theta + 2 \cos^3 \theta = -4 \cos \theta + 6 \cos^3 \theta\).

Thus, \(\frac{dy}{dx} = 6 \cos^5 \theta - 4 \cos^3 \theta\).