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