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I posted an idea which seems to rough to be interesting, so I decided to delete it

Some thoughts on your problem. If $u$ is so smooth that $Du$ has a well defined trace, e.g., $u\in H^s$ for $s>3/2$, then you could flatten the surface by a $C^1$ change of variables and reduce to the case $S=\{x_1=0\}$. Then the assumption on $Du$ becomes $Du=(D_1u,0,...,0)$ which means the derivatives of the trace are zero a.e. on the coordinate plane $S$, and hence $u$ is constant there. This seems to work fine for Sobolev functions and maybe for other functional spaces.

For $BV$ functions I am a bit skeptical; you have a well defined $L^1$ trace for $u$, but in what sense would you formulate your assumption on $Du$ restricted to $S$?

I have not time to check all the details, but maybe you can make something out of this idea. Let $u\in BV$. By a change of variables you can reduce to the case the surface is $x_1=0$, and $u$ has an $L^1$ trace $u(0,x')=f(x')$. Your assumption in this case reduces to $Du=(D_1u,0,...,0)$ a.e. along $x_1=0$, which means $Df=0$. If this is correct, you're done. Necessary background material you can find e.g. in Giusti's book (Minimal Surfaces and BV functions).

I posted an idea which seems to rough to be interesting, so I decided to delete it

I have not time to check all the details, sorry!, but maybe you can make something out of this idea. Let $u\in BV$. By a change of variables you can reduce to the case the surface is $x_1=0$, and $u$ has an $L^1$ trace $u(0,x')=f(x')$. Your assumption in this case reduces to $Du=(D_1u,0,...,0)$ a.e. along $x_1=0$ (first step to check), which means (second step to check) $Df=0$. If this is correct, you're done. Necessary background material you can find e.g. in Giusti's book (Minimal Surfaces and BV functions).

I have not time to check all the details, but maybe you can make something out of this idea. Let $u\in BV$. By a change of variables you can reduce to the case the surface is $x_1=0$, and $u$ has an $L^1$ trace $u(0,x')=f(x')$. Your assumption in this case reduces to $Du=(D_1u,0,...,0)$ a.e. along $x_1=0$, which means $Df=0$. If this is correct, you're done. Necessary background material you can find e.g. in Giusti's book (Minimal Surfaces and BV functions).