Given the curve $$ \vec{\mathbf{r}}(t) = \bigl(t - \tfrac{1}{3}t^3 \bigr) \, \vec{\mathbf{i}} + (4 - t^2) \, \vec{\mathbf{j}}, $$ how can I tell whether it's traveled clockwise or counterclockwise?
Knowing that at the top, it's at $t = 0$ and the intersection is $t = \pm \sqrt{3}$.
It travels clockwise. As $t = 1, x = \dfrac{2}{3}, y = 3 \implies (x,y) = \left(\dfrac{2}{3}, 3\right)$. This point is on the right and in the $1$st quadrant.
While Wang's answer is a correct and direct method, based on considering the position at two different times, another method is to look at the velocity at a single time. Differentiating $$ \vec { \mathbf r } ( t ) = \bigl ( t - \tfrac 1 3 t ^ 3 \bigr ) \, \vec { \mathbf i } + ( 4 - t ^ 2 ) \, \vec { \mathbf j } \text , $$ we get $$ \vec { \mathbf v } ( t ) = \vec { \mathbf r } ' ( t ) = \tfrac { \mathrm d } { \mathrm d t } \vec { \mathbf r } ( t ) = \tfrac { \mathrm d } { \mathrm d t } \bigl ( t - \tfrac 1 3 t ^ 3 \bigr ) \, \vec { \mathbf i } + \tfrac { \mathrm d } { \mathrm d t } ( 4 - t ^ 2 ) \, \vec { \mathbf j } = \cdots \text . $$ Since we already know where $ \vec { \mathbf r } ( 0 ) $, let's use that value of $ t $: $$ \vec { \mathbf v } ( 0 ) = \cdots \text . $$ I leave these for you to calculate; but based on the picture, it had better have a $ \vec { \mathbf j } $-component of $ 0 $ and a positive or negative $ \vec { \mathbf i } $-component. If the $ \vec { \mathbf i } $-component is positive, then it's moving to the right at the top, so clockwise; if the $ \vec { \mathbf i } $-component is negative, then it's moving to the left at the top, so counterclockwise.
