Step 0, draw a diagram. Try to imagine these objects moving around in your head and immediately observe any relationships between these variables. If there's an unknown rate, label it $\frac{d?}{d??}$. Most of the time, these problems are not hard to solve (calculus wise), it's all visualization.
Recall, the rate in the denominator, $d??$ is the "with respect to" portion.
Step 1, list all of our knowns:
1.$$V = \frac{1}{3}\pi r^2h$$
2.$$\frac{dV}{dt} = \frac{-50m^3}{min}$$
3.$$h = 6$$
4.$$h_2 = 5$$
5.$$r = 45$$
Step 2, figure out what we need to solve for, this is your $\frac{d?}{d??}$ In this case, we're looking at the rate of change in height with respect to our radius. Always ask yourself: what is this problem asking me to solve? Write it in terms of $\frac{d?}{d??}$, which in our case is $\frac{dh}{dt}$.
Step 2a, imagine the rate of change you need to solve for. The question asked "at what rate is the water level falling at a height of 5m [in the conic]". So, imagine in your head the water sliding down the conic. Ask yourself, "what's changing?" In this case, the water is sliding down the side of the conic, and as it does so, the radius (and thereby the volume) is changing!
Step 3, identify relationships between variables, keeping in mind our step 2 and step 2a. In this case, it is the relationship between $r$ and $h$ that is so important, so we're going to setup a ratio between $r$ and $h$:
$$\frac{r}{h} = \frac{45}{6}$$
$$ \bf{r = \frac{45h}{6}}$$
Step 4, substitute and simplify as much as reasonably possible.
$$V = \frac{1}{3}\pi r^2h$$
$$V = \frac{1}{3}\pi \left(\bf{\frac{45h}{6}}\right)^2 h$$
$$V = \frac{1}{3}\pi \left(\bf{\frac{15h}{2}}\right)^2 h$$
Step 5, finally, you get to differentiate; however, as this is probably a homework question, I'm going to leave that part to you. One thing to remember is that you don't start plugging numbers in, really until you're almost at the end. Notice here you're subbing in your final $h_2$ value. You're going to get something like:
$$\frac{dV}{dt} = \frac{some number}{some number} \pi \bf{h_2}^2 \bf{\frac{dh}{dt}} $$
Simplify that down and algebraically manipulate it until you get:
$$ some number = \frac{dh}{dt}$$ and that's your answer.
So, how does all this work? (Read: what am I doing?)
So, as you define these relationships that govern this changing thing, what you're really doing is setting up the problem and differentiating to find the rate of change for that problem for some arbitrary instance. After you find the derivative, you substitute in those specific values, in our case your $h_2$, and literally freezing the rate at an exact moment in time to get the rate at which the quantity is changing.
Let me know if this helps.
Best Answer
Since your hemisphere is the lower half of a circle, you can express the radius as the x-coordinate (see image added in edit below),
$$ R^2 = x^2 + y^2 \implies x = \sqrt{R^2 - y^2} $$
now differentiate to find $\frac{dx}{dt}$,
$$ \frac{dx}{dt} = \frac{1}{2} (R^2-y^2)^{-\frac{1}{2}} (-2y) \frac{dy}{dt} = \frac{-y}{\sqrt{R^2-y^2}} \frac{dy}{dt} $$
and you know $y=8$, $R=13$, and $\frac{dy}{dt}=\frac{1}{24\pi}$, so you should be able to plug these values in and get $\frac{dx}{dt} = \frac{-1}{3\pi\sqrt{105}} \approx -0.0103546$
Edit: