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Iosif Pinelis
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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))|} \end{equation} and \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 $(x_u(t),y_u(t))$.

Iosif Pinelis
  • 127.9k
  • 8
  • 107
  • 229