= Solution
{c}
Use the <path integral> with source term
$$
S_{\mathrm{src}}=\int d^4x\,[J^\mu A_\mu+\bar\eta_\alpha\psi_\alpha+\bar\psi_\alpha\eta_\alpha].
$$
The sources $\eta,\bar\eta$ are <Grassmann variables>; all derivatives below are left <functional derivatives>. Because $\eta$ stands to the right of $\bar\psi$, differentiating with respect to it introduces a minus sign. The insertion rules are therefore
$$
\frac1i\frac{\delta}{\delta J^\mu}\longleftrightarrow A_\mu,\qquad
i\frac{\delta}{\delta\eta_\alpha}\longleftrightarrow\bar\psi_\alpha,\qquad
\frac1i\frac{\delta}{\delta\bar\eta_\beta}\longleftrightarrow\psi_\beta.
$$
For example $\delta_L(\bar\psi\eta)/\delta\eta=-\bar\psi$. The order of the odd <functional derivatives> matters: the rightmost derivative acts first, and the displayed product inserts $\bar\psi_\alpha\psi_\beta$ in that order.
Expand $\exp[-ie\int\bar\psi\gamma^\mu\psi A_\mu]$ in its <power series>, replace each field by its source derivative, and pull the derivatives outside the free <path integral>. Re-exponentiating gives
$$
\boxed{Z[J,\eta,\bar\eta]=\exp\left[-ie\int d^4x\left(\frac1i\frac{\delta}{\delta J^\mu(x)}\right)\left(i\frac{\delta}{\delta\eta_\alpha(x)}\right)\gamma^\mu_{\alpha\beta}\left(\frac1i\frac{\delta}{\delta\bar\eta_\beta(x)}\right)\right]Z_0[J,\eta,\bar\eta].}
$$
This identity concerns the unnormalized <generating functional>. Normalized <correlation functions> divide by its zero-source value so that vacuum contributions cancel.
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