Answer
$$
\frac{d y}{d x}= \frac{2t}{3t^2+1}
$$
and at $t=3$, we have
$$
\frac{d y}{d x}= \frac{3}{14}.
$$
Work Step by Step
Since $x=t^3+t$ and $y=t^2-1$, then we have
$$
\frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{y^{\prime}(t)}{x^{\prime}(t)}=\frac{2t}{3t^2+1}
$$
and at $t=3$, we have
$$
\frac{d y}{d x}= \frac{6}{28}=\frac{3}{14}.
$$
