Computation of functional determinant using Feynman diagram 
The above equation is from chapter 9.5 "Functional Quantization of the Spinor Field" of Peskin's and Schroeder's book $($page $305)$. I understand that the initial determinant equal to the final exponent as it is consistent with Eq.$7.98$, but how the intermediate step is got. I think it is related to Eq.$7.96$ but don't know how to explain it further.
 A: This is more a sketch of the calculation than a detailed explanation, and there may be sign errors.
As mentioned at the end of page 304, one can alternatively use Feynman diagrams to evaluate $(9.76)$. To do so one uses two Grassmann-valued sources, say $\eta$ and $\bar{\eta}$, and use the generative functional:
\begin{equation}
Z[\eta,\bar{\eta}]=\int \mathcal{D}\bar{\psi} \mathcal{D}\psi\exp\left[ i\int d^4x \bar{\psi}(i\gamma \cdot D-m)+i\int d^4x (\bar{\eta}\psi+\bar{\psi} \eta)\right]. \tag{1}
\end{equation}
From here it is straightforward to evaluate this expression:
\begin{equation}
Z[\eta,\bar{\eta}] = e^{-e \int d^4x \frac{-i\delta}{\delta \bar{\eta}}\gamma^\mu  A_\mu \frac{i\delta}{\delta \eta}}\det(i\gamma\cdot\partial-m)e^{-i\int d^4x \int d^4y\, \bar{\eta}(x) G(x-y) \eta(y)},\tag{2}
\end{equation}
where $G$ is the Green function of the free spinorial field. It is related to $S_F$, its propagator, by an $i$ factor. Expanding the first exponential in $(2)$ around $\eta=0$ and $\bar{\eta}=0$ we get:
\begin{align}
Z=&\det(i\gamma\cdot\partial-m)\left( 1-i\int d^4x \,\text{tr}\left[(-e \gamma^\mu A_\mu(x))S_F(0)\right] \right. \tag{3.a}
\\
&-\frac{1}{2}\int d^4x d^4y\,\text{tr}\left[(-e \gamma^\mu A_\mu (x))(-e\gamma^\nu A_\nu(y))(iS_F(0))(iS_F(0))\right]\tag{3.b}
\\
&\left.-2\times \frac{1}{2}\int d^4x d^4y\,\text{tr} \left[(-e \gamma^\mu A_\mu (x))(-e \gamma^\nu A_\nu(y))(iS_F(x-y))(iS_F(x-y))\right]+\cdots \right) \tag{3.c}
\end{align}
Normally, the traces appear when you do the calculation with explicit indices. Then, $(3.a)$ corresponds to $1$ plus the first tadpole in the intermediate step of $(9.79)$, $(3.b)$ to the product of the two tadpoles in this intermediate step, and finally $(3.c)$ corresponds to a loop attached with two sources (second loop in the intermediate step).
As I said, I have not checked if my signs are correct, so let me know if there are any errors. Nonetheless, it should be a good sketch of what is going on.
A: *

*The RHS of eq. (9.79) is indeed just the Taylor expansion of the logarithm in eq. (9.78). Note that the  factor $\frac{1}{n}$ from the Taylor expansion matches the (reciprocal) symmetry factor of the corresponding Feynman diagram on the RHS of eq. (9.79).


*The middle expression of eq. (9.79) follows from the RHS via the linked-cluster theorem for connected Feynman diagrams.
