Using differential equation for height of cone 
A cone-shaped reservoir is filled with water through a hole in its
bottom. Water evaporates from the surface and the rate of evaporation
is proportional to the surface area. Use a differential equation to
describe the height of the water in the reservoir.

My solution:
volume of cone-shaped reservoir = volume of water pouring into a cone - surface of evaporated water
$V_{2}=\frac{1}{3}\pi r^{2}h-\pi r^{2}$
I assumed that radius $r$ is function of height $h$ and $V_{2}$ as well as $h$ is function of time $t$, I use a chain rule for radius $r$:
$\frac{dV_{2}}{dt}=\frac{1}{3} 2 \pi r \frac{dr}{dh} \frac{dh}{dt}- 2 \pi r \frac{dr}{dh}$
What is equal to:
$\frac{dV_{2}}{dt}=\frac{1}{3} 2 \pi r \frac{dr}{dt} - 2\pi r \frac{dr}{dh}$
But I do not know how to continue. Is my solution correct?
 A: HINT:
Aim to get a product rather than a sum:
$$ V =\dfrac{ \pi r^2 h}{3}- k \pi r^2= \pi r^2(\dfrac{h}{3}-k) $$
Perform log differentiation
$$\dfrac{dV}{V}=2\dfrac{dr}{r}+  \dfrac{dh/3}{\dfrac{h}{3}-k}$$
Plug in for $V$ from first equation and simplify, bearing in mind we can afterwards divide each of
$$ (dV,dr,dh)\; \text{  by}  \;dt $$
to obtain differential coefficients from differentials.
A: First of all, we can write the volume as
$$V(t)=\dfrac{\pi\tan(\theta)^2}{3}h(t)^3$$ where we have used the fact that $r(h)=h\tan(\theta)$ and $\theta$ is the angle of the one.
Clearly, we have that
$$\dot{V}(t)=\dot{h}(t)h(t)^2\pi\tan(\theta)^2.$$
Now, it is said that the evaporation rate is proportional to the surface, which is $\pi r(t)^2=\pi h(t)^2\tan(\theta)^2$. So, let $\alpha>0$ be this proportionality constant. Now, since this is an evaporation rate, it acts at the level of the volume variation. Therefore, the volume changes according to
$$\dot{V}(t)=-\alpha \pi h(t)^2\tan(\theta)^2$$
where the negative signs comes from the fact that evaporation decreases the volume.
Substitution yields
$$\dot{h}(t)h(t)^2\pi\tan(\theta)^2=-\alpha \pi h(t)^2\tan(\theta)^2.$$
Assuming that $h(t)>0$, then this simplifies to
$$\dot{h}(t)=-\alpha,$$
which means that evaporation decreases the height at a constant rate, until the height reaches zero.
