Let us say we have a function $f(2x)$, and let us say $u = 2x$.
It should be clear by the chain rule that $$\frac{d}{dx}f(2x) = 2f'(2x) = 2f'(u) = 2\frac{df}{du}$$ Now $$\frac{df}{du} = \frac{df}{dx}\frac{dx}{du} = \frac12\,\frac{df}{dx}$$ So the expression above simply becomes $$\frac{df}{dx} = \frac{d}{dx}f(x)$$ Thus we have seemingly proven that $f(2x) = f(x)$ without defining $f$?
I do not understand what is going on here but I am clearly confusing something about derivative rules.
On
I will write out a way to think of this situation with clear notation.
Take a function $f:\mathbb R \to \mathbb R$ and denote the doubling function $u:\mathbb R\to \mathbb R$ given by $u(x)=2x$, whose derivative is $u'=2$. With this information define a new function $F:\mathbb R\to\mathbb R$ by $F=f\circ u$. Then $$F'=(f\circ u)'=(f'\circ u)\cdot u'= 2(f'\circ u).$$ Evaluating at $x$, you get $$F'(x)=2f'(u(x))=2f'(2x).$$ Now with the $d/d$ notation, probably the best way to write this is $$\frac{dF}{dx}(x)=2\frac{df}{du}(2x)$$ or something like that. The problem is that everyone has their own rationale for how to use the $d/d$ notation.
Edit: I want to clarify something about the use of language. A function $f:\mathbb R\to\mathbb R$ is a function of one variable and hence it can only be differentiated with respect to one thing: its argument. The result of differentiating $f$ is another function $f':\mathbb R\to\mathbb R$ defined by $$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}.$$
Now, the phrase "the derivative of $f$ with respect to $u$" refers to this function $f'$ evaluated at $u$. But it is still the same function $f'$, just evaluated at another point $u$. That is $$f'(u)=\lim_{h\to 0}\frac{f(u+h)-f(u)}{h}.$$
If $u = 2x$, then by the chain rule, $$ \frac{\mathrm d}{\mathrm dx}f(2x) = f'(2x) \frac{\mathrm d}{\mathrm dx}(2x) = 2f'(2x) = 2f'(u) = 2 \frac{\mathrm d}{\mathrm du}f(u).\tag1 $$
Another way to write the same application of the chain rule, but substituting $u = 2x$ earlier, is $$ \frac{\mathrm d}{\mathrm dx}f(2x) = \frac{\mathrm d}{\mathrm dx}f(u) = \left(\frac{\mathrm d}{\mathrm du}f(u)\right) \frac{\mathrm du}{\mathrm dx} = 2f'(u) = 2 \frac{\mathrm d}{\mathrm du}f(u). $$
Now you can also apply the chain rule like this:
$$ \frac{\mathrm d}{\mathrm du}f(u) = \left(\frac{\mathrm d}{\mathrm dx}f(u)\right) \frac{\mathrm dx}{\mathrm du} = \frac12 \frac{\mathrm d}{\mathrm dx}f(2x). \tag2 $$
If you use Equation $(2)$ to substitute for $\frac{\mathrm d}{\mathrm du}f(u)$ on the far right-hand side of Equation $(1)$, you will find that $$ \frac{\mathrm d}{\mathrm dx}f(2x) = 2 \left(\frac12 \frac{\mathrm d}{\mathrm dx}f(2x)\right) = \frac{\mathrm d}{\mathrm dx}f(2x), $$ as you might expect.
What you cannot legitimately do is to write $$ \frac{\mathrm d}{\mathrm du}f(u) \stackrel?= \left(\frac{\mathrm d}{\mathrm dx}f(\color{red}{x})\right) \frac{\mathrm dx}{\mathrm du}. $$
In general, $$ \frac{\mathrm d}{\mathrm du}f(u) = \frac{\mathrm d}{\mathrm dx}f(x) \neq \frac{\mathrm d}{\mathrm dx}f(u), \text{ where $u = 2x$}. $$
As mentioned already by others, the fatal flaw in your argument has to do with the Leibniz notation $\frac{\mathrm df}{\mathrm dx}$. You can get away with this notation when $f$ is actually a variable whose value is determined by $x$. But if $f$ is literally the name of a function, you need to provide a value of that function in the Leibniz notation, for example, $\frac{\mathrm df(x)}{\mathrm dx}$ or $\frac{\mathrm df(u)}{\mathrm dx}$.
What you did in the question was to write $\frac{\mathrm df}{\mathrm dx}$ in a place where only $\frac{\mathrm df(u)}{\mathrm dx}$ makes sense, but later you reinterpreted the same value of $\frac{\mathrm df}{\mathrm dx}$ as $\frac{\mathrm df(x)}{\mathrm dx}$. In other words, you misused the notation $\frac{\mathrm df}{\mathrm dx}$ to silently replace $f(u)$ with $f(x)$. No wonder you derived the incorrect conclusion then that $f(u) = f(x)$.