Equality of functions
The actual problem is that what you wrote there are not the functions, but just two terms varying by a certain parameter.
In order to call those a function, you have to define the domain, which would look like this:
$$f: \mathbb{R}\setminus\{2, 4\}\rightarrow \mathbb{R}, x \mapsto f(x) := \frac{(x-2)(x-3)}{(x-2)(x-4)}\\ g: \mathbb{R}\rightarrow \mathbb{R}, x \mapsto g(x) := \frac{(x-3)}{(x-4)}\\$$ Now, $f\ne g$, but it equals the restriction of g to the domain of f: $f = g|_{\mathbb{R}\setminus \{2, 4\}}$
EDIT: Let me explain a bit more about how to compare terms and functions regarding equality.
Functions are defined as relations with certain properties (uniqueness & definedness of the argument): $f: A\rightarrow B, x\mapsto f(x) :\Leftrightarrow f := \{(x, y)\in A\times B|y = f(x)\}$, which has to hold $\forall x\in A\exists!y\in B: (x, y)\in f$ as the defining property of a function.
If two functions are equal is therefore just a question of comparing two sets. It holds if both contain exactly the same elements $(x, f(x))$. Thus, they must have the same domain.
When comparing two expressions regarding equality, one has to be careful as well: Let $f(x), g(x)$ be expressions. $f(x) = g(x)$ would not be a logical statement, because We do not know what x is — therefore, we have to give an explicit x or quantorize it: $$\forall x: f(x) = g(x)$$ Because a quantorization without limiting x to a certain set (let's say $\mathbb{R}$) would only make sense if we would concern f(x) = g(x) = x, which would allow x to be something else than a number, you have to give a domain in some way as well. For instance, let $f(x) := \frac{x-1}{x-1}, g(x) := 1$: $\forall x\in M: f(x)=g(x)$ would hold true for M = $\mathbb{R}\setminus \{1\}$, but not for $\mathbb{R}$, because there is no $f(x)$ at $x=1$.
They are not. You can only say they assume the same values in the intersection of their domains, that is: $$f|_{D(f) \cap D(g)} = g|_{D(f) \cap D(g)}.$$
This answer assumes the definition: two functions $f:A \to B$ and $g:C \to D$ are equal if: $A = B$, $C = D$ and $f(x) = g(x)$ for all $x \in A = C$.