Let $X(t)$ be a martingale (in continuous time), and for each real $u$, let $Y_u(t)=g(u,X(t))$ for some infinitely-differentiable $g:\mathbb{R}^2\to\mathbb{R}$, and assume that $Y_u(t)$ is a martingale as well.
Does it hold, for each natural $k>0$ and each real $v$, that $$ \frac{d^k}{du^k} Y_u(t) \restriction_{u=v} $$ is a martingale?
If this does not hold in general, does it hold for the case where $g(x,y)=e^{xy}$?
If the function $g$ is sufficiently "nice", then
$$\left(\frac{d^k}{du^k} Y_u(t) \right)_{t \geq 0}$$
is in fact a martingale.
Using induction, it suffices to consider the case $k=1$. Recall that a process $(Z_t)_{t \geq 0}$ is a martingale with respect to a filtration $(\mathcal{F}_t)_{t \geq 0}$ if, and only if,
$$\int_F Z_s \, d\mathbb{P} = \int_F Z_t \, d\mathbb{P} \tag{1}$$
for any $s < t$ and $F \in \mathcal{F}_s$. Now fix $s<t$ and $F \in \mathcal{F}_s$. Applying $(1)$ to the martingale $Y_u(t) = g(u,X_t)$, we find
$$\int_F g(u,X_{s}) \, d\mathbb{P} = \int_F g(u,X_t) \, d\mathbb{P}$$
for any $u \in \mathbb{R}$. Hence,
$$\frac{1}{h} \left( \int_F (g(u+h,X_s)-g(u,X_s)) \, d\mathbb{P} \right) = \frac{1}{h} \left( \int_F (g(u+h,X_t)-g(u,X_t)) \, d\mathbb{P} \right). \tag{2}$$
If $g$ is sufficiently nice (e.g. if $g$ has bounded derivatives or the derivates have a particular form, see the example below), then we can apply the dominated convergence theorem and obtain
$$\int_F \frac{d}{du} g(u,X_s) \, d\mathbb{P} = \int_F \frac{d}{du} g(u,X_t) \, d\mathbb{P}.$$
This finishes the proof.
Example If we consider $g(u,x) = e^{u \cdot x}$, then $$\frac{d}{du} g(u,x) = x \cdot e^{ux}.$$ By assumption, we know that $\mathbb{E}e^{uX_t} <\infty$ and from the Cauchy-Schwarz inequality it is not difficult to see that we may apply the dominated convergence theorem in $(2)$.
Applying this argumentation to a Brownian motion $(B_t)_{t \geq 0}$, one can show that the processes
$$B_t^2 -t, \qquad B_t^3-3t B_t, \qquad B_t^4-6B_t^2+3t^2, \qquad \ldots$$
are martingales.