For fun, I like to liven-up the "black box"/machine view of a function by putting a monkey into the box. (I got pretty good at chalkboard-sketching a monkey that looked a little bit like Curious George, but with a tail.)
Give the Function Monkey an input and he'll cheerfully give you an output. The Function Monkey is smart enough to read and follow rules, and make computations, but he's not qualified to make decisions: his rules must provide for exactly one output for a given input. (Never let a Monkey choose!)
You can continue the metaphor by discussing the monkey's "domain" as the inputs he understands (what he can control); giving him an input outside his domain just confuses and frightens him ... or, depending upon the nature of the audience, kills him. (What? You gave the Reciprocal Monkey a Zero? You killed the Function Monkey!) Of course, it's probably more appropriate to say that the Function Monkey simply ignores such inputs, but students seem to like the drama. (As warnings go, "Don't kill the Function Monkey!" gets more attention than "Don't bore the Function Monkey!")
The Function Monkey comes in handy later when you start graphing functions: imagine that the x-axis is covered with coconuts (one coconut per "x" value). The Function Monkey strolls along the axis, picks up a "x" coconut, computes the associated "y" value (because that's what he does), and then throws the coconut up (or down) the appropriate height above (or below) the axis, where it magically sticks (or hovers or whatever). So, if you ever want to plot a function, just "Be a Function Monkey and throw some coconuts around". (Warning: Students may insist that that's not a coconut the Monkey is throwing.)
Further on, you can make the case that we're smarter than monkeys (at least, we should strive to be): We don't always have to mindlessly plot points to know what the graph of an equation looks like; we can sometimes anticipate the outcome by studying the equation. This motivates manipulating an equation to tease out clues about the shape of its graph, explaining, for instance, our interest in the slope-intercept form of a line equation (and the almost-never-taught intercept-intercept form, which I personally like a lot), the special forms of conic section equations (which aren't all functions, of course), and all that stuff related to translations and scaling.
Parametric equations can be presented as a way to let the Function Monkey plot elaborate curves ... both in the plane and in space (and beyond).
All in all, I find that the Function Monkey can make the course material more engaging without dumbing it down; he provides a fun way to interpret the definitions and behaviors of functions, not a way to avoid them. Now, is the Function Monkey too cutesy for a College Algebra class? My high school students loved him, even at the Calculus level. One former student told me that he would often invoke the Function Monkey when tutoring his college peers. If it's clear to the students that the instructor isn't trying to patronize them, the Function Monkey may prove quite helpful.
The first and last relations are not transitive. Can you see why? You're right that in the first case, since $(a, b), (b, c) \in R$, we'd need $(a,c)\in R$ for the relation to be transitive. $(a, c)\notin R$, so the relation is not transitive.
What's missing in the last relation? We have $(b, c)$ and $(c, b)$ in the relation, so $b\mapsto c \mapsto b$, but $(b, b)\notin R$. Hence, the relation cannot be transitive.
For transitivity, if $(a, b)$ and $(b, c)$ are in the relation, then we must have $(a, c)$ in the relation. But $a, b, c$ need not be distinct. So the problem in the last case isn't about no relation between $a, c$. $a$ is related only to itself, and no other element is related to $a$, so since $(a, a)\in R$, $a$ is "off the hook" here. The problem in the last case is that if it were transitive, then it would be true that $\Big((b, c)\in R \land (c, b)\in R\Big) \implies (b, b) \in R$. It's not, so the relation is not transitive.
Best Answer
In fewer symbols, the relation you are trying to show is an equivalence relation is that two functions $f$ and $g$ are equivalent if their derivatives are the same. The definition of an equivalence relation is a relation which is symmetric, reflexive, and transitive, so all you need to do is to prove the relation is each of these three things.
Again in fewer symbols (and expanding the definitions), here is what you need to prove: (some might come across as too obvious for proof; that's because the equality is already an equivalence relation)
(Symmetric). Let $f$ and $g$ be differentiable functions. If $f$'s derivative is equal to $g$'s derivative then $g$'s derivative is equal to $f$'s derivative.
(Reflexive) Let $f$ be a differentiable function. Then $f$'s derivative is equal to itself.
(Transitive) Let $f,g,h$ be differentiable. Then if $f$'s derivative is equal to $g$'s and $g$'s is equal to $h$'s, then $f$'s is equal to $h$'s.
Once this is done, one may entertain the relation's equivalence classes. One way to do this is to find a representative for each. This is basically the way to the answer: if $f$ and $g$ are in the same equivalence class, then they have the same derivative. By the Mean Value Theorem, we can deduce that $f$ and $g$ are different by at most a constant (i.e., $f(x) = g(x) + c$ for some real number $c$). Conversely, if $f$ and $g$ are two differentiable functions different by a constant, then they are equivalent (easy exercise). Thus, there is one equivalence class for each function, modulo addition by a constant.