Answer
$$
s^{3}-4st+2t^{3}-5t=0
$$
The velocity $\frac{ds}{dt}$ is given by:
$$
\begin{aligned}
\frac{d s}{d t} &=\frac{4 s-6 t^{2}+5}{3 s^{2}-4 t}
\end{aligned}
$$
Work Step by Step
$$
s^{3}-4st+2t^{3}-5t=0
$$
Now, we can calculate the velocity $\frac{ds}{dt}$ by implicit differentiation, use the product rule as follows:
$$
\begin{aligned}
\frac{d }{d t}( s^{3}-4 s t+2 t^{3}-5 t)&=\frac{d}{d t}(0) \\
3 s^{2} \frac{d s}{d t}-\left(4 t \frac{d s}{d t}+4 s\right)+6 t^{2}-5 &=0 \\
3 s^{2} \frac{d s}{d t}-4 t \frac{d s}{d t}-4 s+6 t^{2}-5=0 \\
\frac{d s}{d t}\left(3 s^{2}-4 t\right) &=4 s-6 t^{2}+5 \\
\frac{d s}{d t} &=\frac{4 s-6 t^{2}+5}{3 s^{2}-4 t}
\end{aligned}
$$
