$xy=4$
$a)$ Find $\dfrac{\mathrm dy}{\mathrm dt}$ when $x=8$, Given $\dfrac{\mathrm dx}{\mathrm dt} = 10$
$b)$ Find $\dfrac{\mathrm dx}{\mathrm dt}$ when $x=1$, Given $\dfrac{\mathrm dy}{\mathrm dt} = -6$
for problem $a$, I got up to $\dfrac{\mathrm dy}{\mathrm dt} = \dfrac{-y}{x} \dfrac{\mathrm dx}{\mathrm dt}$, but I'm not sure what to do next. Not sure if its even right.
For example: $xy=4 \\ x=8 \text{ so we have } 8y=4 \\ \text{ and so } y=\frac{1}{2} \\ \frac{dy}{dt}=\frac{-y}{x} \cdot \frac{dx}{dt}=-\frac{\frac{1}{2}}{8} \cdot 10 \text{ when } x=8 \\ \text{ I will leave the other one as an exercise to you }$