= Solution
<Differentiation> of $x=\rho\cos\phi$, $y=\rho\sin\phi$ gives
$$
h_\rho=1,\quad e_\rho=(\cos\phi,\sin\phi,0),\qquad
h_\phi=\rho,\quad e_\phi=(-\sin\phi,\cos\phi,0).
$$
The two radial face areas are $\rho_0\Delta\phi\Delta z$ and $(\rho_0+\Delta\rho)\Delta\phi\Delta z$. Each constant-$\phi$ face has area $\Delta\rho\Delta z$, and each constant-$z$ face has area
$$
\frac12\{(\rho_0+\Delta\rho)^2-\rho_0^2\}\Delta\phi.
$$
The volume is this last expression times $\Delta z$.
Only the radial faces contribute to the flux of $F(\rho)e_\rho$. Dividing their difference by the volume and taking $\Delta\rho\to0$ gives
$$
\frac1\rho\frac{d}{d\rho}\{\rho F(\rho)\}
$$
at $\rho=\rho_0$.
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