This problem was posed previously (5 years ago here), but the user's question was about terminology - not how to solve the problem. I would like to know how to solve the problem.
Question: A tank for storing water is conical in shape, with maximum radius 20 m and maximum depth 15 m. The large, circular end of the tank points vertically upwards and is open to the air. It is known that the volume of water in the tank will decrease at a rate proportional to the area of the water's surface, due to evaporation. (1) Show that the water-level in the tank decreases at a constant rate. (2) If the tank originally stores $1000 ~m^3$ of water and the water-level drops $1~cm/day$, how long will it be before the tank is empty? [Hint: A cone of base radius r and height h has a volume $\frac {1}{3}\pi r^2h]$.
My attempt (at part (1)):
This is my revised version after discussion with @DreamConspiracy.
Ultimately we would like to find something like $\frac {dh}{dt} = k$.
We are told that the change in the volume is proportional to the surface area of the water. The surface area is described by the area of a circle, $A = \pi r^2$. Therefore,
$$\frac {dV}{dA} = -k\pi r^2$$
$k$ is the constant of proportionality. The relationship is negative because the water volume is decreasing. The volume of the cone is
$$V = \frac {1}{3} \pi r^2h$$
or, substituting $A$ for $\pi r^2$, as suggested by @DreamConspiracy,
$$V = \frac {1}{3} Ah$$
Now differentiate:
$$\frac {dV}{dt} = \frac {1}{3}(\frac {dA}{dt}h + A\frac{dh}{dt})$$
So what is $\frac {dA}{dt}$?
$$A = \pi r^2$$ $$\frac {dA}{dt} = 2 \pi r \frac{dr}{dt}$$
What is $\frac {dr}{dt}$?
$$tan~\alpha = \frac {r}{h}$$ $$h~tan~\alpha = r$$ $$\frac {dh}{dt}tan~\alpha = \frac {dr}{dt}~~~(tan~\alpha~is~constant)$$
What is $h$? It is $\frac {r}{tan~\alpha}$.
We were given the fact that $\frac {dV}{dt} = -kA$. We have all of the information we need. Substituting the respective expressions for $\frac {dA}{dt}$, $\frac {dr}{dt}$ and $h$ and rearranging, we get
$$\frac {dV}{dt} = \frac {1}{3}(\frac {dA}{dt}h + A\frac{dh}{dt})$$ $$\frac {dV}{dt} = \frac {1}{3}(\underbrace{2 \pi r \frac {dr}{dt}}_{\frac {dA}{dt}}h+A \frac{dh}{dt})$$ $$\frac {dV}{dt} = \frac {1}{3}(2 \pi r \underbrace{\frac {dh}{dt}tan~\alpha}_{\frac{dr}{dt}}~h+A \frac{dh}{dt})$$ $$\frac {dV}{dt} = \frac {1}{3}(2 \pi r \frac {dh}{dt}tan~\alpha \underbrace{\frac {r}{tan~\alpha}}_{h}+A \frac{dh}{dt})$$ $$\frac {dV}{dt} = \frac {1}{3}\underbrace{\frac {dh}{dt}}_{\text{factor}}(2 \pi r ~tan~\alpha\frac {r}{tan~\alpha}+A)$$ $$\frac {dV}{dt} = \frac {1}{3}\frac {dh}{dt}(\underbrace{\frac {2 \pi r^2 ~tan~\alpha}{tan~\alpha}}_{\text{rearrange}}+A)$$ $$\frac {dV}{dt} = \frac {1}{3}\frac {dh}{dt}(2 \pi r^2 +A)$$ $$\frac {dV}{dt} = \frac {1}{3}\frac {dh}{dt}(2A +A)$$ $$\frac {dV}{dt} = \frac {1}{3}\frac {dh}{dt}(3A)$$ $$\frac {dV}{dt} = \frac {dh}{dt}\cdot A$$
Substituting $-kA$ for $\frac {dV}{dt}$ yields $$-kA = \frac {dh}{dt}\cdot A$$ $$\frac {-kA}{A} = \frac {dh}{dt}$$ $$\frac {dh}{dt} = -k$$
@DreamConspiracy's solution below is "cleaner" because he sets out immediately to express $r$ in terms of $h$ to simplify the volume equation. $(r = h~tan~\theta~ = h \cdot \frac {20}{15} = h \cdot \frac {4}{3})$.
This is correct. In particular, $V=\frac 13 Ah$. But then $$\frac{\partial V}{\partial A} = \frac h3 \neq \frac{2\pi rh}3.$$
Edit: There is a second more subtle mistake in your method. You write $\frac{\partial V}{\partial A}=-k\pi r^2$. However, this is not quite correct. Looking at the sentence from which you drew this conclusion, we can see $$\text{Rate at which volume of water in tank decreases}\propto \text{Surface area}$$ $$\frac{dV}{dt}=-kA$$
With problems like this, it is also helpful to begin by "making a plan". That is, you want to find $\frac{dh}{dt}$. So you begin by writing: $$\frac{dh}{dt}=\frac{dh}{dV}\frac{dV}{dt}$$. This makes sense because you already know what $\frac{dV}{dt}$ is, and $\frac{dh}{dV}$ does not depend on $t$. Now we can see that $$V=\frac{Ah}3$$ $$V=\frac{\pi \left( \frac{4h}{3}\right)^2h}{3}$$ $$V=\frac{16\pi h^3}{27}$$ $$\frac{dV}{dh}=\frac{16\pi h^2}{9}$$ $$\frac{dV}{dh}=\frac{16 \pi \left( \frac 34r\right)^2}9$$ $$\frac{dh}{dV}=\frac 1A.$$ Now we can plug into our equation above to obtain $$\begin{aligned} \frac{dh}{dt}&=\frac{dh}{dV}\frac{dV}{dt} \\ &=\frac 1A (-kA) \\ &=-k \end{aligned}$$