 Solution

For $0<\theta <\pi$, the chain rule for a parametric curve givesAt $\theta =0$ the quotient of parameter derivatives is formally $0/0$, but the simplified expression has limit $16$. Thus 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 $y=8x^4-8x^2+1$.