Answer
Horizontal tangent at $(0,-3)$
Vertical tangents at $(2,-2)$ and $(-2,-2)$
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Work Step by Step
$\displaystyle \frac{dy}{dx}=\frac{dy/dt}{dx/dt}$
The tangent is horizontal for $\displaystyle \frac{dy}{dt}=0$
$y=t^{2}-3$
$\displaystyle \quad \frac{dy}{dt}=2t,$
$2t=0\Rightarrow t=0$
When $t=0, (x,y)=(0,-3)$
The tangent is vertical for $\displaystyle \frac{dx}{dt}=0$
$x=t^{3}-3t$
$\displaystyle \quad \frac{dx}{dt}=3t^{2}-3=3(t+1)(t-1),$
$t=-1$ or $1$
When t=1, $(x,y)=(2,-2)$, and
when t=$-1, (x,y)=(-2,-2).$
