Let $f:\mathbb{Q}_p\to \mathbb{R} $ with $f(x)=(1/|x|_p)^2$ if $x\neq 0$ and $f(0)=0$ . How to prove that f is differentiable on $\mathbb{Q}_p$.
I am a beginner in this field of p-adic numbers $$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}h=\lim_{h\to 0}\frac{(p^{\nu_p(x+h)})^2-(p^{\nu_p(x)})^2}h$$ How to continuous? because I did not understand this solution
As said in the comments, for the definition of a derivative to make sense, we need to view this as a function $f: \mathbb Q_p \rightarrow \mathbb Q_p$. Then
$$f'(x)=\displaystyle \lim_{h\to 0}\frac{f(x+h)-f(x)}h=\begin{cases}\displaystyle \lim_{h\to 0}\dfrac{p^{2\nu_p(x+h)}-p^{2\nu_p(x)}}h \text{ if } x \neq 0 \\ \displaystyle\lim_{h\to 0}\dfrac{p^{2\nu_p(h)}}h = \lim_{h\to 0}\dfrac{1}{\lvert h\rvert_p^2 h} \text{ if } x=0 \end{cases}.$$
Note that "$h \to 0$" here is meant in the $p$-adic sense, i.e. $\nu_p(h) \to \infty$, i.e. $\lvert h \rvert_p \to 0$.
Now in the first case, a fundamental principle of ultrametrics says that
$$\nu_p(x+h) = \nu_p(x) \text{ for all } h \text{ with } \lvert h \rvert_p < \lvert x \rvert_p$$
so that (since $x \neq 0$) we have $\displaystyle \lim_{h\to 0}\dfrac{p^{2\nu_p(x+h)}-p^{2\nu_p(x)}}h = \lim_{h \to 0}\dfrac{0}{h} =0$.
(And the same would be true without the square: The absolute value function is locally constant away from $0$, hence has local difference quotient $0$.)
In the second case, we have $\lvert \dfrac{1}{\lvert h\rvert_p^2 h} \rvert_p = \dfrac{\lvert h \rvert_p^2}{\lvert h \rvert_p} = \lvert h \rvert_p$ and hence $\displaystyle\lim_{h \to 0} \dfrac{1}{\lvert h\rvert_p^2 h} =0$ as well. Note that without the square in the function, we would get $f'(0) =1$ here.