I have the PDE $$u_t(t) - \Delta f(u(t)) = 0$$ in $H^{-1}(\Omega)$ where $f$ is a nonlinear function.

Define $F(s) = \int_0^s f(s)$. Note that if $u_t(t) \in L^2(\Omega)$, $$\frac{d}{dt}F(u(t)) = f(u(t))u_t(t).$$

We could test the PDE with $f(u(t))$ (i.e multiply by it and integrate) and if $u_t \in L^2(\Omega)$, we can use the above identity to get $$\int_{\Omega}F(u(T)) - \int_{\Omega}F(u_0) + \int_0^T \int_{\Omega} |\nabla f(u(t))|^2 =0$$ and then we can throw away the first term on the LHS since $F$ is positive (a fact for the particular problem). This gives us a bound on $ \nabla f(u)$ in $L^2(0,T;L^2)$.

This is a standard trick. What can I do if I only have $u_t \in H^{-1}(\Omega)$? What is the trick in this case?