When applying Faddeev and Popov method (am using Pesking and Shcroeder as reference), we use the identity: $$1=\int \mathcal{D}\alpha(x)\delta(G(A^\alpha)) \det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right)$$ to write $$ \int \mathcal{D}Ae^{iS[A]}=\det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right)\int\mathcal{D}\alpha(x)\int\mathcal{D}Ae^{iS[A]}\delta (G(A^\alpha))$$ When we use the Lorentz gauge, we obtain: $$G(A)=\partial^\mu A_\mu+\frac{1}{e}\partial^2\alpha(x)$$

My questions is: how do we obtain the following: $$ \det\left(\frac{\delta G(A^\alpha)}{\delta\alpha}\right) = \det\left(\frac{1}{e}\partial^2\right) $$

I am also confused about how can we obtain an operator by taking a functional derivative, if someone could give an intuitive explanation of this it would be highly appreciated!