Answer
$(0,0)$ and $(2,-4)$ are point of horizontal tangent.
$(-2,-2)$ and $(2,-4)$ are points of vertical tangents.
Work Step by Step
\[x=t^3-3t\;\;,\;\;y=t^3-3t^2\]
Differentiate $x$ with respect to $t$
\[\frac{dx}{dt}=3t^2-3\]
Differentiate $y$ with respect to $t$
\[\frac{dy}{dt}=3t^2-6t\]
\[\frac{dy}{dx}=\frac{\displaystyle\frac{dy}{dt}}{\displaystyle\frac{dx}{dt}}=\frac{3t^2-6t}{3t^2-3}=\frac{t^2-2t}{t^2-1}\]
The curve will have horizontal tangent if $\displaystyle\frac{dy}{dx}=0$, i.e., $\displaystyle\frac{dy}{dt}=0$ but $\displaystyle\frac{dx}{dt}\neq 0$
\[\frac{dy}{dt}=0\Rightarrow t^2-2t=0\Rightarrow t=0,2\]
If $t=0 \Rightarrow x=(0)^3-3(0)=0\;,\;y=(0)^3-3(0)^2=0$
If $t=2 \Rightarrow x=(2)^3-3(2)=2\;,\;y=2^3-3(2)^2=-4$
$(0,0)$ and $(2,-4)$ are point of horizontal tangent.
For vertical tangent if $\displaystyle\frac{dx}{dt}=0$ but $\displaystyle\frac{dy}{dt}\neq 0$
\[\frac{dx}{dt}=0\Rightarrow 3t^2-3=0\Rightarrow t=\pm 1\]
If $t=1 \Rightarrow x=1^3-3(1)=-2\;,\; y=1^3-3(1)^2=-2$
If $t=-1\Rightarrow x=(-1)^3-3(-1)=2\;,\;y=(-1)^3-3(-1)^2=-4$
So $(-2,-2)$ and $(2,-4)$ are points of vertical tangents.