Although I have asked a similar question here, here, I find that I don't totally understand it, so I arrange my new ideas to this post.
Begin with Berezin integral: $$\left(\prod_i \int d \theta_i^* d \theta_i\right) e^{-\theta_i^* B_{i j} \theta_j}=\operatorname{det} B \tag{9.69}. $$
Now I want to calculate: $$\int \mathcal{D} \bar{\psi} \mathcal{D} \psi \exp \left[i \int d^4 x \bar{\psi}(i \not \partial-m) \psi\right], \tag{A}$$ the book said that $A=\text{det}(i\not\partial-m)$, but that's not clear for me.
First, we need to note that the indices $i,j$ in (9.69) is the discretization of all space time, and $i$, $j$ are independent. But now our situation is $\bar{\psi}(x)(i \not \partial-m)_x \psi(x)$, we only have one set of space time points (in Hilbert space), so I want to expand operator $(i\not\partial-m)_x$ in matrix representation.
Then, if we define the Green function of $(i\not\partial-m)_x$: $$(i\not\partial-m)_x S_F(x-y)=i\delta^4(x-y), \tag{B}$$ we can represent (see here) $$(i\not\partial-m)_x \psi(x)=\int d^4 y (-i S_F(x-y)^{-1})\psi(y), \tag{C} $$ Now we can write A as: $$ \begin{aligned} A&=\int \mathcal{D} \bar{\psi} \mathcal{D} \psi \exp \left[i \int d^4 x d^4 y\bar{\psi}(x)[-i S_F(x-y)^{-1}] \psi(y)\right]\\ & = \Pi_i \int d\bar{\psi}_i d\psi_i \exp \left[- \bar{\psi}_i[- S_F(x-y)^{-1}]_{ij} \psi_j\right], \end{aligned} \tag{D}$$ where in the second line we use spacetime discretization, and $i$, $j$ sum separately inside the exponent.
So from D and (9.69), it seems that $$\int \mathcal{D} \bar{\psi} \mathcal{D} \psi \exp \left[i \int d^4 x \bar{\psi}(i \not \partial-m) \psi\right]=\text{det}(- S_F(x-y)^{-1}). \tag{E}$$
So would $$\text{det}(- S_F(x-y)^{-1})=\text{det}(i\not\partial-m)~?\tag{F}$$ Or where is my problem?
