Answer
The points where the tangent line is horizontal are:
$c(0)=(0,48)$,
$c(\sqrt{6})=(2 \sqrt{6},12)$,
$c(-\sqrt{6})=(-2 \sqrt{6},12)$.
The points where the tangent line is vertical are:
$c(\sqrt{4/3})=(-16/(3 \sqrt{3}),304/9)$,
$c(-\sqrt{4/3})=(16/(3 \sqrt{3}),304/9)$.
Work Step by Step
Using Eq. (8) the slope of the tangent line is
$\frac{dy}{dx}=\frac{y'(t)}{x'(t)}=\frac{4 t^3-24 t}{3 t^2-4}$.
The tangent line is horizontal if $\frac{dy}{dx}=0$. So, we solve the equation $\frac{4 t^3-24 t}{3 t^2-4}=0$.
$4t (t^2-6)=0$,
$4t (t-\sqrt{6})(t+\sqrt{6})=0$.
The solutions are $t=0$, $t=\sqrt{6}$ or $t=-\sqrt{6}$. The corresponding points where the tangent line is horizontal are:
$c(0)=(0,48)$,
$c(\sqrt{6})=(2 \sqrt{6},12)$,
$c(-\sqrt{6})=(-2 \sqrt{6},12)$.
The tangent line is vertical if $\frac{dy}{dx}$ is infinite. This occurs if $3t^2 - 4=0$. The solutions are $t=\sqrt{4/3}$ or $t=-\sqrt{4/3}$. The corresponding points where the tangent line is vertical are:
$c(\sqrt{4/3})=(-16/(3 \sqrt{3}),304/9)$,
$c(-\sqrt{4/3})=(16/(3 \sqrt{3}),304/9)$.
