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