Answer
$$ \frac{d y}{d x}= \frac{2}{3t} $$
and at $t=-4$, we have $$ \frac{d y}{d x}= -\frac{1}{6}. $$
Work Step by Step
Since $x=t^3$ and $y=t^2-1$ then we have
$$ \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{y^{\prime}(t)}{x^{\prime}(t)}=\frac{2t}{3t^2}=\frac{2}{3t} $$
and at $t=-4$, we have $$ \frac{d y}{d x}= \frac{2}{3(-4)}=-\frac{2}{12}=-\frac{1}{6}. $$
