Let $f: \mathbb R^2 \rightarrow \mathbb R$ be a smooth strictly convex function with unique minimum at $0$ such that all level sets $A_x:=\left\{z ; f(z) \le x \right\}$ are compact. Imagine something like $f(z)=\Vert z \Vert^2.$
Define the integral function
$$F(x):=\int_{A_x} g(z) dz$$
where $g$ is as smooth as you like.
Question: Is there a way to analytically determine an expression for $F''(x)$?
Let $B_u:=\{z\colon f(z)=u\}$ for $u>u_0:=\min f$. Let $[0,2\pi)\ni t\mapsto(x_u(t),y_u(t))$ be any smooth parametrization of $B_u$, so that $B_u=\{(x_u(t),y_u(t))\colon t\in[0,2\pi)\}$.
(For instance, one may take $(x_u(t),y_u(t))=(\rho_u(t)\cos t,\rho_u(t)\sin t)$, where $\rho_u(t):=f_t^{-1}(u)$ and $f_t^{-1}\colon(u_0,\infty)\to(0,\infty)$ is the function inverse to the function $f_t\colon(0,\infty)\to(u_0,\infty)$ defined by the formula $f_t(r):=f(r\cos t,r\sin t)$.)
Then \begin{equation} F'(u)=\int_0^{2\pi}dt\,\sqrt{x'_u(t)^2+y'_u(t)^2}\frac{g(x_u(t),y_u(t))}{|(\nabla f)(x_u(t),y_u(t))|}, \tag{$\ast$} \end{equation} and hence \begin{equation} F''(u)=\int_0^{2\pi}dt\,\frac d{du}\Big(\sqrt{x'_u(t)^2+y'_u(t)^2}\frac{g(x_u(t),y_u(t))}{|(\nabla f)(x_u(t),y_u(t))|}\Big). \end{equation}
Here, $dt\,\sqrt{x'_u(t)^2+y'_u(t)^2}$ is the infinitesimal length element of the curve $B_u$, and $\frac{du}{|(\nabla f)(x_u(t),y_u(t))|}$ is the infinitesimal distance between the curves $B_u$ and $B_{u+du}$ near the point $(x_u(t),y_u(t))$.
We can verify formula $(\ast)$ for e.g. $f(x,y)=x^2+y^4$ and $g(x,y)=y^2$, in which case we have the following closed-form expressions: for $u>0$ \begin{equation} F(u)=\frac{2 \sqrt{\pi } \Gamma \left(\frac{7}{4}\right)}{3 \Gamma \left(\frac{9}{4}\right)}\,u^{5/4} ,\quad F'(u)=\frac{5 \sqrt{\pi } \Gamma \left(\frac{7}{4}\right)}{6 \Gamma \left(\frac{9}{4}\right)}\,u^{1/4}, \end{equation} and the latter expression coincides with the integral in $(\ast)$, with the parametrization $(x_u(t),y_u(t))=(u^{1/2}\cos t,u^{1/4}(\sin t)^{[1/2]})$ of $B_u$, where $w^{[1/2]}:=|w|^{1/2}\,\text{sign}\, w$.
