Answer
$y=x$ and $y=-x$ are both tangent to the curve at $(0,0)$.
Work Step by Step
Given
$$
x=\cos t,\ \ \ \ y=\sin t \cos t $$
Since
\begin{aligned}
\frac{dy}{dt}&=-\sin ^2 t+\cos ^2 t\\
&=\cos 2 t \\
\frac{dx}{dt}&= -\sin t
\end{aligned}
At $(x,y)=(0,0)$, we have
\begin{aligned}
\cos t&=0\ \ \to \ \ t=\frac{\pi}{2}, \frac{3\pi}{2},\ \cdots \\
\end{aligned}
At $t=\pi/2$, we have
$$d x / d t=-1\ \ \to \ \ d y / d t=-1,\ \ \ d y / d x=1$$
At $t=3\pi/2$, we have
$$ d x / d t=1\ \to \ d y / d t=-1\ \ , \ \ d y / d x=-1.$$
It follows that $y=x$ and $y=-x$ are both tangent to the curve at $(0,0)$.
