I've recently come across this functional:
$F[f] = \int \frac{|\nabla f|^2}{f} dr$ (in 3D space, if that's relevant)
and am interested in taking the functional derivative $\frac{\delta F}{\delta f}$. I've seen examples for when the integrand of $F$ is $|f'|^2$, where an integration by parts and suitable boundary conditions on $f$ gives the desired result, however it is not clear to me how this approach would generalize to my example, or in general where the integrand involves a derivative of $f$.
Thank you for tips!
Edit: this is my current approach, that I hope is correct:
$ \begin{align} F[f+\delta f] &= \int{\frac{|\nabla (f+\delta f)|^2}{f+\delta f} dr}\\ &\approx\int{|\nabla (f+\delta f)|^2 \cdot \frac{1}{f} \left(1-\frac{\delta f}{f}+...\right) \;dr}\\ & = \int{\left[\frac{|\nabla f|^2}{f} - \frac{2}{f}|\nabla f|\cdot|\nabla(\delta f)| - \frac{|\nabla f|^2}{f^2}\delta f + O(|\delta f|^2)\right]dr} \end{align}$
Then, I need to integrate the term with $\nabla(\delta f)$ by parts (with suitable boundary conditions) to move the gradient operation onto the other term, collect the terms of order $\Delta f$, and identify the result multiplying $\delta f$ as $\frac{\delta F}{\delta f}$.
I would use the product rule. Since $$\delta |\nabla f|^2 = 2 \nabla f \nabla (\delta f)$$ and $$\delta f^{-1} = - f^{-2} (\delta f)$$ it follows that $$\delta |\nabla f|^2 f^{-1} = 2 \nabla f \nabla (\delta f) f^{-1} - |\nabla f|^2 f^{-2} (\delta f)$$ Integrate both terms, the first one by parts (I don't know your boundary conditions, so I just assume the boundary term vanishes): $$ \int \left[-2\nabla\cdot (f^{-1}\nabla f ) - |\nabla f|^2 f^{-2} \right] (\delta f)$$ And use the product and chain rules for $\nabla\cdot (f^{-1}\nabla f )$: $$ \nabla\cdot (f^{-1}\nabla f ) = \nabla (f^{-1})\cdot \nabla f + f^{-1} \Delta f = -f^{-2} |\nabla f|^2 + f^{-1} \Delta f $$ So the end result is $$ \int \left[2 f^{-2}|\nabla f|^2 -2 f^{-1}\Delta f - |\nabla f|^2 f^{-2} \right] (\delta f) = \int \left[f^{-2}|\nabla f|^2 -2 f^{-1}\Delta f \right] (\delta f) $$