The answer is no.

Indeed, if a martingale is a.s. everywhere differentiable, then its [quadratic variation][1] is a.s $0$. So, by the [Burkholder--Davis--Gundy inequality][2], the martingale is a.s. constant.

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**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][2], 
\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][3], (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][4]) 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.  



  [1]: https://en.wikipedia.org/wiki/Quadratic_variation#Definition
  [2]: https://en.wikipedia.org/wiki/Quadratic_variation#Martingales
  [3]: https://www.ams.org/journals/tran/2011-363-08/S0002-9947-2011-05209-8/S0002-9947-2011-05209-8.pdf
  [4]: https://mathoverflow.net/a/398042/36721