Answer
Horizontal tangent at $(2,-4)$ and $(0,0)$
Vertical tangents at $(2,-4)$ 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 $\displaystyle \frac{dy}{dt}=0$
$y=t^{3}-3t^{2}$
$\displaystyle \quad \frac{dy}{dt}=3t^{2}-6t=3t(t-2),$
$\displaystyle \frac{dy}{dt}=0\Rightarrow t=0$ or $t=2.$
When $t=2, \quad(x,y)=(2,-4)$
When $t=0, (x,y)=(0,0)$
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$=1,\ \quad(x,y)=(-2,-2)$, and
when t=$-1, \quad (x,y)=(2,-4).$
