Might be a stupid question, but I was thinking about this today and wanted to clear something up. I always thought of f(x) as having different algebraic properties than f×x, and it came as a surprise to me when I noticed that you can modify the equation algebraically, treating $f(x)$ as $f×x$, without necessarily butchering the original function.
Most rudimentarily:
$f(x) = x$
$\frac {f(x)}x = \frac xx$
$f(1) = 1$
I was able to get the answer by simply dividing x out of the function $f(x)$ as if it were being multiplied by another variable.
Another example:
$f(x) = x+2$
$f(x) × \frac2x=\frac{2x}x+\frac4x$
$f(2) = 2+\frac4x$
This one's a little different, as you'd need to plug the value within f that has replaced x into the equation, but
$f(x) = 2+\frac4x$ (the equation we ended up with)
$2+\frac4{(2)} = 4 = f(2)$
$f(x) = x+2$
$(2)+2=4 = f(2)$
Yes, I realize that if I were to plug anything into x that was not the constant in the parentheses, the new derived function would not work. Regardless, I thought it was interesting that function notation and algebraic notation have such a distinctive intersection (for lack of better phrasing). The only difference between $f(x)$ and $f×x$ is that the equation on the opposite end of $f(x)$ remains consistent regardless of what number you plug in in place of x, whereas the equation accompanying $f×x$ changes based on what you want to substitute for x. Is this just another mathematical coincidence, of sorts?
Apologies if this is fallaciously thought out, or poorly worded. I don't know math in and out and this thought is rather fresh.
It sounds like you're interested in functions $f$ with the property that for all numbers $x$ and $y$, we have $f(y)=f(x)\frac yx$. This is equivalent to $\frac{f(x)}x=\frac{f(y)}y$, i.e. $\frac{f(x)}x$ is a constant $c$ that doesn't depend on $x$. So the only functions with this property are of the form $f(x)=cx$. This makes sense in hindsight, since you're trying to treat the "$f$" as a multiplied constant.
In your second example, you "cheated" by plugging 2 in again for the remaining occurrence of $x$. Functions for which this works satisfy $f(y)=f(x)\frac yx[x\to y]$, where the $[x\to y]$ indicates that we replace occurrences of $x$ with $y$. But this clearly true for all functions because $f(x)\frac yx[x\to y]=f(y)\frac yy=f(y)$. In other words, you're multiplying $f(x)$ by $\frac yx$ and then plugging in $y$ for $x$ so that the $f(x)$ becomes $f(y)$ and the $\frac yx$ becomes $1$.