Let $\mathbb F$ be a field with characteristic $2$, and with a non-trivial absolute value. (The simplest example is $\mathbb F_2(T)$, the field of formal rational expressions with coefficients in $\mathbb F_2$. The absolute value of a polynomial is $|p(T)|=2^{-n}\in\mathbb R$, where $p(T)$ has a factor of $T^n$ but not $T^{n+1}$.)
Let $f:\mathbb F\to\mathbb F$ be a function. Limits and derivatives are defined as usual: $\lim_{x\to a}f(x)=b$ means that for any real (or rational) $\varepsilon>0$ there exists $\delta>0$ such that, for all $x\in\mathbb F$ where $0<|x-a|<\delta$, $|f(x)-b|<\varepsilon$. And
$$f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}.$$
Is it possible that $f''(a)$ exists in $\mathbb F\setminus\{0\}$?
If $f=g\cdot h$ where $g$ and $h$ are twice differentiable, the product rule gives
$$f''=g''h+2g'h'+gh''$$ $$=g''h+gh''.$$
If $f=g\circ h$, the chain rule gives
$$f''=(g''\circ h)\cdot h'^2+(g'\circ h)\cdot h''.$$
Any polynomial function $p(x)=\sum a_nx^n$ has second derivative $p''(x)=\sum a_nn(n-1)x^{n-2}=0$ because $n(n-1)$ is always even. The derivatives of the reciprocal $f(x)=1/x$ are $f'(x)=-1/x^2$ and $f''(x)=2/x^3=0$. It follows from the above properties that any rational function $f(x)=p(x)/q(x)$ also has $f''(x)=0$. The square root function satisfies $f(x)^2-x=0$; differentiating this gives $-1=0$, so it must not be differentiable. I'm not sure about general algebraic functions.
I thought of using the symmetric second derivative formula, but I don't think it's valid:
$$f''(a)\overset?=\lim_{x\to0}\frac{f(a+x)-2f(a)+f(a-x)}{x^2}=\lim_{x\to0}\frac{0}{x^2}$$
Well, let's apply the definition directly and see what happens:
$$f''(a)=\lim_{x\to0}\frac{f'(a+x)-f'(a)}{x}$$ $$=\lim_{x\to0}\frac{\lim_{y\to0}\frac{f(a+x+y)-f(a+x)}{y}-\lim_{y\to0}\frac{f(a+y)-f(a)}{y}}{x}$$ $$=\lim_{x\to0}\lim_{y\to0}\frac{f(a+x+y)-f(a+x)-f(a+y)+f(a)}{xy}$$
We can't take $y=x$ here to get $f''=0$, because it's not a single limit $(x,y)\to(0,0)$ but two nested limits.
An equivalent description of differentiability is that there's some continuous function $h$ with $h(0)=0$ such that
$$f(a+x)=f(a)+xf'(a)+xh(x).$$
So the second derivative is
$$f''(a)=\lim_{x\to0}\lim_{y\to0}\frac{(x+y)h(x+y)-xh(x)-yh(y)}{xy}$$ $$=\lim_{x\to0}\lim_{y\to0}\left(\frac{h(x+y)-h(x)}{y}+\frac{h(x+y)-h(y)}{x}\right)$$ $$=\lim_{x\to0}\left(\lim_{y\to0}\frac{h(x+y)-h(x)}{y}+\frac{h(x)}{x}\right)$$ $$=\lim_{x\to0}\left(h'(x)+\frac{h(x)}{x}\right).$$
Note that all of these limits must exist, in order for $f''(a)$ to exist. But we can't simplify this further, since $h'$ may be discontinuous or undefined at $0$.
We may even consider the special case where $f$ has a $2$nd-order Peano derivative: there's some continuous function $g$ with $g(0)=0$ such that
$$f(a+x)=f(a)+xf'(a)+x^2f^\ddagger(a)+x^2g(x).$$
Then the second derivative is
$$f''(a)=2f^\ddagger(a)+\lim_{x\to0}xg'(x).$$
Is it possible that this limit exists in $\mathbb F\setminus\{0\}$?
