Rate of change for water in inverted cone 
Water flows into an inverted cone of height $h$ and radius $r$ at a constant rate of $k$ cm$^3$s$^{-1}$. Find the rate of change of the curved surface area of the cone in contact with the water in terms of $h$ and $k$ when the cone is filled to $\frac18$ of its volume.

Taking the volume of water in the cone to be $v$ and the curved surface area of the cone in contact with the water to be $a$,
$${dv \over dt} = k\\
{da \over dt} = {dv \over dt} \cdot {da \over dv}\\
{da \over dt} = k \cdot {dh \over dv} \cdot {da \over dh}$$
However, I am unable to utilise the $\frac18$ volume given in the question. The answer given $k \pi \sqrt{r^2+h^2}$ cm$^2$s$^{-1}$.
Can someone explain how to solve this?
 A: I don't think the answer you offered is correct, at least in terms of how the problem is stated. The units are wrong. $k\pi \sqrt{r^{2}+h^{2}}$ has units of $cm^{4}/s$, which wouldn't work for the time derivative of an area. I obtained a different, unit consistent answer when I attempted the problem myself.
You're on the right track. I would take care to treat $r$ and $h$ as fixed values and assign variables to the depth and surface radius of the water currently in the take (e.g. $H$ and $R$ respectively). The volume of the water in the tank is given by $\frac{\pi}{3}R^{2}H$ and since we are told that the tank is $\frac{1}{8}$ full, we can write
$$\frac{\pi}{3}R^{2}H = \frac{1}{8}\frac{\pi}{3}r^2h$$
Also, note that we can enforce
$$\frac{R}{H} = \frac{r}{h}$$
From these two equations it is possible to write $H$ as a function of $h$ and $R$ as a function of $r$. If we proceed with the notation change I have suggested (and let $A$ represent the curved surface area of the current water volume), your chain rule relation becomes
$$\frac{dA}{dt} = \frac{dA}{dH}\frac{dH}{dV}\frac{dV}{dt}$$
Once you solve for $\frac{dA}{dt}$ in terms of $R$ and $H$, you can use the relations above to find $\frac{dA}{dt}$ in terms of $r$ and $h$.
