let $\Omega\subset\mathbb{R}^n$ be a bounded domain and $f\in C^1(\bar\Omega \times (0,g(x)),[0,1])$ and $f(x,\cdot)$ increasing and $g(x)\in\mathbb{R}$ continuous (maybe better, Lipschitz?). I want to consider for $u\in H^1(\Omega)$, $u(x)\in[0,g(x)]$ for almost every $x$ the composition $f(x,u(x))$. However since $f$ is not $C^1$ on $\bar\Omega \times [0,g(x)]$ one does not a priori know whether such a function is in $H^1(\Omega)$. However if $f$ had this better regularity ($C^1$ on $\bar\Omega \times [0,g(x)]$) case, we would get the chainrule $$ \nabla f(x,u)= f_x(x,u) + \partial_u f(x,u)\nabla u$$ simply by approximation.
However, for the case without so much regularity this won't work as Tomas pointed out. To this end we need to cut off the function $f$ at the levels say $1-\varepsilon$ and $\varepsilon$.
The "Stampacchia" lemma (i.e. $D(u-\varepsilon)_+ =Du \cdot 1_{u>\varepsilon}$) only cuts off on hyperplanes. But in this problem the cut-off level in space since $f$ depends on $x$. Any suggestion?
Cutting-off at the levels $1-\varepsilon$ and $\varepsilon$ I expect the derivative to be
$$\nabla f_\varepsilon(x,u)= 1_{\{\varepsilon<f(x,u)<1-\varepsilon\}}(f_x(x,u) +\partial_uf(x,u) \nabla u)$$
Potentially, we can write $f_\varepsilon(x,u)=(g\circ f)(x,u)$ with $g(a)=(\max\{1-\varepsilon,a\}-\varepsilon)_+ +\varepsilon$.
Without $x$-dependence of $f$ I can prove this with the Chainrule for Sobolev-functions since $f$ is Lipschitz and maps from $\mathbb{R}to\mathbb{R}$. Then in the chainrule the term $f_x$ vanishes. In the case with $x$-dependence the problem is that the set $\{f^{-1}(1-\varepsilon)\}$ is not anymore a single point but a curve/surface.
Check the second comment of Tomas and my answer for a good example.