Answer
Horizontal tangents at $\left(\frac{1}{2},-1\right)$, $\left(-\frac{1}{2},1\right)$
No vertical tangents
Work Step by Step
$x$ = $\theta\cos\theta$
$y$ = $\cos3\theta$
$\frac{dy}{d\theta}$ = $-3\sin3\theta$
$\frac{dy}{dθ}$ = $0$
$-3\sin3\theta$ = $0$
$3\theta$ = $0, \pi, 2\pi, 3\pi$
$\theta$ = $0$, $\frac{\pi}{3}$, $\frac{2\pi}{3}$, $\pi$
$(x,y)$ = $(1,1)$, $\left(\frac{1}{2},-1\right)$, $\left(-\frac{1}{2},1\right)$, $(-1,-1)$
$\frac{dx}{d\theta}$ = $-\sin\theta$
$\frac{dx}{d\theta}$ = $0$
$-\sin\theta$ = $0$
$\theta$ = $0, \pi$
$(x,y)$ = $(1,1)$, $(-1,-1)$
The curve has a horizontal tangent when $\frac{dy}{dx}=0$, that is, when $\frac{dy}{d\theta}=0$ and $\frac{dx}{d\theta}\not=0$.
The slope when $\theta$ = $0$
$\lim\limits_{\theta \to 0}{\frac{dy}{dx}}$ = $\lim\limits_{\theta \to 0}{\frac{-3\sin3\theta}{-\sin\theta}}$ = $\lim\limits_{\theta \to 0}{\frac{-9\cos3\theta}{-\cos\theta}}$ = $9$
The curve has horizontal tangents at $\left(\frac{1}{2},-1\right)$, $\left(-\frac{1}{2},1\right)$
The curve has a vertical tangent when $\frac{dx}{d\theta}=0$.
There are no vertical tangents.
