= Solution
For $0<\theta<\pi$, the <chain rule> for a <parametric curve> gives
$$
\boxed{\frac{dy}{dx}=\frac{-4\sin4\theta}{-\sin\theta}=16\cos\theta\cos2\theta.}
$$
At $\theta=0$ the quotient of parameter derivatives is formally $0/0$, but the simplified expression has limit $16$. Thus \b[the limiting tangent at $(x,y)=(1,1)$ has gradient 16]. This is an endpoint limit of the given open parameter interval, or the ordinary tangent of the smooth polynomial extension $y=8x^4-8x^2+1$.
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