The answer is no.

Indeed, if a martingale is a.s. everywhere differentiable, then its quadratic variation is a.s $0$. So, by the Burkholder–Davis–Gundy inequality, the martingale is a.s. constant.

The answer is no.

Indeed, if a martingale is a.s. everywhere differentiable, then its quadratic variation is a.s $0$. So, by the Burkholder--Davis--Gundy inequality, the martingale is a.s. constant.

Details: Suppose that $X:=(X_t)_{t\in[0,1]}$ is an almost surely (a.s.) everywhere differentiable martingale. Replacing $X_t$ by $X_t-X_0$, without loss of generality let us assume that $X_0=0$. Take any real $a>0$ and consider the bounded martingale $X^a:=(X_t^a)_{t\ge0}$, where $X_t^a:=X_{\min(t,T_a)}$ and $T_a:=\inf\{t\in[0,1]\colon|X_t|=a\}$, with $\inf\emptyset:=\infty$.

For any real-valued function $f$ on $[0,1]$, define its quadratic variation by the formula $$[f]:=\limsup\sum_{j=1}^n(f(t_j)-f(t_{j-1}))^2,$$ where the $\limsup$ is taken over all "partitions" $0=t_0<\cdots<t_n=1$ of $[0,1]$ as $\max_{1\le j\le n}(t_j-t_{j-1})\to0$.

By the Burkholder--Davis--Gundy inequality, \begin{equation} c\,E([X^a]^{1/2})\le EM^a\le C\,E([X^a]^{1/2}), \tag{BDG} \end{equation} where $c$ and $C$ are universal positive real constants and $M^a:=\max_{t\in[0,1]}|X_t^a|$.

By the first one of inequalities (BDG), $[X^a]<\infty$ a.s. So, using (i) Corollary 23 in this paper, (ii) the remark on line 3 of page 4228 of the same paper that $\mu_f=0$ iff $f\in V_2^0$, (iii) the obvious identity $\mu_f([0,1])=[f]$, and (iv) the definition of $V_2$ as the set of all functions $f$ with $[f]<\infty$, we conclude (as in this answer) that $[X^a]=0$ a.s. Therefore, by the second one of inequalities (BDG), $M^a=0$ a.s. for each $a$ and hence $X=0$ a.s., as desired.