The construction of an $\epsilon$-$\delta$ proof is usually exactly opposite its presentation. This should be a good example. Given $\epsilon$, you need to be able to produce some $\delta$ such that for all $x$ with $|x-1|<\delta$, $\frac{2 + 4x}{3} - 2|<\epsilon$.
How do you do this? You (usually) need some formula to produce $\delta$ in terms of $\epsilon$. So start with the expression you need,
$$\bigg|\frac{2 + 4x}{3} - 2\bigg|<\epsilon,$$
and start solving for $x$ in terms of $\epsilon$.
I'll let you work out the details here; it's a simple algebraic exercise. At the end, you'll get something like
$$\bigg|x - 1\bigg| < \frac{3}{4}\epsilon.$$
Ah-hah! Now you have produced a constraint on $|x-1|$ in terms of $\epsilon$. If I give you $\epsilon$, you can pick $\delta < \frac{3}{4}\epsilon$ and, as you can quickly verify, this $\delta$ will satisfy the $\epsilon$ bound. In fact, it has to --- that's how you made it.
The take-home lesson here, again, is that the way you build the proof is backwards. Start with what you want and backsolve for what you need. Then when you write the proof out, you know how to choose $\delta$, so you can quickly verify that $\lim_{x\to 1}\frac{2 + 4x}{3} = 2.$
Q1. It means exactly what it says. :-) How much does one variable change, with respect to (that is, in comparison to) another variable? For instance, if $y = 3x$, then the derivative of $y$, with respect to $x$, is $3$, because for every unit change in $x$, you get a three-unit change in $y$.
Of course, that's not at all complicated, because the function is linear. With a quadratic equation, such as $y = x^2+1$, the derivative changes, because the function is curved, and its slope changes. Its derivative is, in fact, $2x$. That means that at $x = 1$, an infinitesimally small unit change in $x$ gives a $2x = 2$ unit change in $y$. This ratio is only exact right at $x = 1$; for example, at $x = 2$, the ratio is $2x = 4$.
This expression is the limit of the ratio $\frac{\Delta y}{\Delta x}$, the change in $y$ over the change in $x$, over a small but positive interval. The limit as that interval shrinks to zero is $\frac{dy}{dx}$.
Q2. You will rarely see, at this stage, $\frac{d}{dx}$ by itself. It will be a unary prefix operator, operating on an expression such as $x^2+1$. For instance, we might write
$$
\frac{d}{dx} \left(x^2+1\right) = 2x
$$
It just means the derivative of the expression that follows.
Q3. This is an unusual formulation. Ostensibly, though, it would mean the derivative of the operand with respect to $f(x)$, which you can obtain using the chain rule:
$$
\frac{dx}{df(x)} = \frac{\frac{dx}{dx}}{\frac{df(x)}{dx}} = \frac{1}{f'(x)}
$$
and
$$
\frac{d}{df(x)} g(x) = \frac{\frac{dg(x)}{dx}}{\frac{df(x)}{dx}}
= \frac{g'(x)}{f'(x)}
$$
Best Answer
For functions of one variable, I have never seen a problem, and wouldn't hesitate to treat them as fractions (multiplicatively). However, suppose you have $F(x,y)$ which implicitly defines a function $y=f(x)$, then
$\dfrac{dy}{dx} = -\dfrac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$
If you just straight cancel as fractions, you'd get the wrong sign.
Edit: I just thought; there is a problem with the notation used which makes this error possible. The $\partial F$'s are different! One is given constant $x$ and the other constant $y$, hence they shouldn't necessarily cancel as they do in fractions. I guess it's rather pretty how they do manage to cancel to give a $-1$, but this particular case as just one instance, it's entirely possible for one symbol to represent different things in the same expression, so one would have to be much more careful about cancelling terms.