Everything in the picture is a function of time AB & BC

A related rates problem always has the same shape. Two or more quantities are tied together by a geometric or algebraic relationship, and all of them are changing as time passes. Differentiating the relationship with respect to $t$ produces an equation relating their rates.

The one habit that makes the rest work: write every varying quantity as a function of time, out loud, before differentiating. A circle's area and radius are not $A$ and $r$; they are $A(t)$ and $r(t)$, tied by

$$A(t) = \pi\left[r(t)\right]^{2}.$$

Written that way, differentiating both sides with respect to $t$ is just the chain rule from Topic 3.1:

$$\frac{dA}{dt} = 2\pi r \frac{dr}{dt}.$$

The factor $\frac{dr}{dt}$ is not decoration. It is the derivative of the inside function, and dropping it produces $\frac{dA}{dt} = 2\pi r$, which is not even dimensionally sensible: the left side is an area per unit time and the right side is a length.

The rule generalizes with no exceptions. Differentiating with respect to $t$:

1. $V = \frac{4}{3}\pi r^{3} \;\Rightarrow\; \dfrac{dV}{dt} = 4\pi r^{2}\dfrac{dr}{dt}$.
2. $x^{2} + y^{2} = 169 \;\Rightarrow\; 2x\dfrac{dx}{dt} + 2y\dfrac{dy}{dt} = 0$.
3. $xy = 12 \;\Rightarrow\; \dfrac{dx}{dt}\,y + x\dfrac{dy}{dt} = 0$, by the product rule, since both factors move.

When a relationship holds three varying quantities, all three rate factors appear. For a cylinder whose radius and height both change,

$$V = \pi r^{2}h \;\Rightarrow\; \frac{dV}{dt} = \pi\left(2r\frac{dr}{dt}h + r^{2}\frac{dh}{dt}\right),$$

which is the product rule applied to $r^2 \cdot h$, with a chain-rule factor on each. Attaching $\frac{dr}{dt}$ to one term and forgetting $\frac{dh}{dt}$ on the other is the most common way this goes wrong, and it silently assumes the height is frozen.

A genuine constant, by contrast, contributes nothing. A ladder's length really is fixed, so in $x^2 + y^2 = 169$ the right side differentiates to 0.

Deciding what is fixed and what is moving is a modeling question, and it is worth spending real time on. Two categories that get confused:

1. Genuinely constant: the length of a rigid ladder, the full dimensions of a tank, the height of a lamppost, the fixed 15 miles per hour of a car described as travelling at a constant speed.
2. Only momentarily equal to a number: the water's depth in the tank right now, the radius of the oil slick at this instant, the distance between two cars at 2 pm.

Anything in the second group is a function of $t$ that happens to have a known value at one moment. It carries a rate factor and it must not be replaced by its number before the differentiation. Topic 4.5 makes that trap explicit; here it is enough to notice which letters belong in which group.

The constraint is the other half of this. A cone-shaped tank whose full radius is 4 and full height 12 forces the water's surface radius and depth to satisfy $\frac{r}{h} = \frac{4}{12} = \frac{1}{3}$ at every instant, by similar triangles. Using it to write $r = \frac{h}{3}$ turns a two-variable relationship into one variable:

$$V = \frac{1}{3}\pi r^{2}h = \frac{1}{3}\pi\left(\frac{h}{3}\right)^{2}h = \frac{\pi h^{3}}{27}.$$

Now only $\frac{dh}{dt}$ appears. Skipping the constraint leaves an equation with two unknown rates and one equation, which cannot be solved.

Five steps, and the last one is where 4.5 picks up.

1. Draw it, and label every length with a letter, not a number, if it is changing.
2. Name the rates. Which derivative is given, and which is wanted? Write both down in $\frac{d\,?}{dt}$ form with their signs: something shrinking gets a minus.
3. Find the relationship tying the two quantities together. Similar triangles, the Pythagorean relation, a volume or area formula, or a stated model.
4. Use any constraint to eliminate variables you have no rate for.
5. Differentiate both sides with respect to $t$, attaching a rate factor to every varying letter. Only then substitute the instant's numbers.

Step 2 deserves the extra half minute. A great many wrong answers are a correctly solved problem that answered for $\frac{dx}{dt}$ when $\frac{dy}{dt}$ was asked for, or that treated the given rate as the unknown.