I'm interested in whether I can make $x\mapsto x+\dfrac{2^{\nu_2(x)}}{3}$ differentiable at $0$ in $\Bbb Q_2$.
For the avoidance of doubt I'm defining $2^{\nu_2(0)}=0$
My attempt is:
$\displaystyle f'(x)=\lim_{x\to a}\dfrac{f(x)-f(a)}{x-a}$
Then as a first example I'm taking $x=0$ and looking at the sequence $a=2^n:n\to\infty$
$\displaystyle f'(x)=\lim_{n\to \infty}\dfrac{2^n+2^n/3}{2^n}=4/3$
Of course there are two routes to take here:
$2^n/2^n=1$, or
$2^n/2^n=1$ provided $2^n\neq0$, because at $2^n=0$ we have a division by zero.
Is there a serious problem with the $2^n/2^n=1$ answer?
Here's a concern I have with the $2^n/2^n=1$ provided $n\neq0$ answer: I am pretty certain the function topologically conjugates to another function $f$, which exchanges the point at zero with another point in the space, whose derivative is well-defined. That seems problematic to me.
To sum up comments and get this off the unanswered list:
First of all, for the purposes of this question, the behaviour of the function $g(x) = x + \dfrac{2^{\nu_2(x)}}{3}$ at $0$ is easily translated to that of $f(x) = 3\cdot g(x) - 3x = 2^{\nu_2(x)} = \frac{1}{\lvert x \rvert_2}$ at $0$.
This $f$, when viewed as a (locally constant!) function $(\mathbb Q_2\setminus \{0\}, \lvert\cdot\rvert_2) \rightarrow (\mathbb Q_2, \lvert\cdot\rvert_2)$, can be made continuous at $0$ by setting $f(0)=0$. But this extension is not differentiable at $0$:
In fact, any sequence $(x_n)_n$ in $\mathbb Q_2\setminus \{0\}$ which $2$-adically goes to $0$ can be written $x_n = u_n 2^{y_n}$, with $(y_n)_n$ a sequence of natural numbers going to infinity, and $(u_n)_n$ a sequence of $2$-adic units $\mathbb Z_2^\times$. And conversely, any combination of such a $(y_n)_n$ and $(u_n)_n$ gives a $2$-adic sequence going to $0$. But
$$\dfrac{f(x_n)-f(0)}{x_n-0} = \frac{1}{u_n}$$
i.e. for any $2$-adic unit $u$ you want (say, $17$), you can find a sequence going to $0$ such that the differential quotient in question goes to that number. And that's already achievable by looking at the sequence $x_n = \frac{1}{u}2^n$ with constant $u_n := u$.
(This is user coiso' point in a comment where their $a$ is our $u$. Since the limit of the differential quotient of $g$, if it exists for a certain sequence, is $1$ plus that of $f$ divided by $3$, this also retrieves Jyrki's examples of getting $1+1/3$ for the sequence $x_n = 2^n$ i.e. $u=1$ but $1-1/3$ for the sequence $x_n = -2^n$ i.e. $u=-1$.)
For most sequences, this will not even converge.