To be specific, let us use the notations in W. Metzner et al., Rev. Mod. Phys. 84,299 (2012). The generating functional $G[\eta, \bar\eta]$ is given by [Eq. 4] \begin{align} G[\eta, \bar\eta] = - \ln \int D\psi D\bar\psi e^{-S[\psi,\bar\psi]} e^{(\bar\eta,\psi)+(\bar\psi,\eta)}. \end{align}
I want to know what is the functional derivative $\partial G /\partial \eta$? I have two guesses.
First, Below Eq. 9, the authors showed that \begin{align} \frac{\partial G }{ \partial \eta} = \bar\psi \end{align} and in Eq. 19 the authors have used a trick where $V[\psi,\bar\psi]$ is replaced by $V[\partial_{\bar\eta}, \partial_{\eta}]$. This clearly shows that the functional derivative of the generating functional $G[\eta,\bar\eta]$ with respect to the source field $\eta$ is the conjugate field $\bar\psi$.
My second guess is that
\begin{align}
\frac{\partial{G}}{\partial \eta} =
-\frac{\int D\psi D\bar\psi [-\bar\psi] e^{-S[\psi,\bar\psi]}e^{(\bar\eta,\psi)+(\bar\psi, \eta)}}{\int D\psi D\bar\psi e^{-S[\psi,\bar\psi]}e^{(\bar\eta,\psi)+(\bar\psi, \eta)}}.
\end{align}
This is the expectation value of the field operator with finit source.
In the zero souce case, this gives the expectation value
\begin{align}
\frac{\partial{G}}{\partial \eta}\Big{|}_{\eta = \bar\eta = 0} =
-\frac{\int D\psi D\bar\psi [-\bar\psi] e^{-S[\psi,\bar\psi]}}{\int D\psi D\bar\psi e^{-S[\psi,\bar\psi]}}
= \langle \bar\psi \rangle
\end{align}
By taking the second-order derivative, we have
\begin{align}
\frac{\partial^2{G}}{\partial\bar\eta\partial \eta}\Big{|}_{\eta = \bar\eta = 0}
= -\langle \bar\psi \rangle \langle \psi \rangle + \langle \psi \bar\psi \rangle
= \langle \psi \bar\psi \rangle_{c}
\end{align}
This coincides with the Eq. 7 with $m = 1$.
I do not know how to reconcile these two cases.
So what is the functional derivative of the generating functional $G[\eta, \bar\eta]$ with respect to the source field $\eta$? Is it the conjugate field $\bar\psi$ or its expectation value $\langle \bar\psi\rangle$ with finit source?
