Let $\Phi : \mathbb{R}^n \to \mathbb{R}^k$ be a smooth map, with $n>k$ and let $f:\mathbb{R}^n \to \mathbb{R}$ be a smooth function. Let $J\Phi(x)$ be the Jacobian of $\Phi$ in $x$: it is defined as $\sqrt{\det(D\Phi(x)\cdot D\Phi(x)^t)}$ where $D\Phi(x)$ is the differential of $\Phi$ in $x$. Assume that $f\cdot(J\Phi)^{-1}$ is integrable. By the coarea formula, for almost all points $z \in \mathbb{R}^k$, $$p(z)= \int_{\Phi^{-1}(\{z\})} f(x)(J\Phi(x))^{-1}d\mathcal{H}^{n-k}(x)$$ is well defined and defines a measurable function on $\mathbb{R}^k$. Is this function smooth as well ? If not, under what conditions is $p$ smooth? (For instance, $\mathbb{R}^n$ and $\mathbb{R}^k$ can be replaced by smooth compact manifolds if needed)
Edit: definition of the Jacobian.
Since you are dividing by $J\Phi$, I assume that you are assuming that $J\Phi>0$ everywhere. But then if you fix $s_0$ and $x_0$ you can apply the implicit function theorem to say that the equation $\Phi(x)-s=0$ in a neighborhood of $(x_0,s_0)$ can be written as the graph of a smooth function. In particular this function is smooth in the variable $s$. So locally you have a parametrization of the family of manifolds $\Phi(x)=s$ which depends smoothly on $s$. Hence, you can differentiate under the integral sign and get regularity of $p$. Since this is only true in a neighborhood of $x_0$, you will need a partition of unity.