Two-point Green for Free Dirac Fields I am trying to compute the $2$-point Green function $\tau_2(x,y)$ for free Dirac fields. The corresponding formula for $\tau_2(x,y)$ is given by
$$\tau_2(x,y) = -\frac{\delta^2}{\delta\eta_x \delta \bar{\eta}_y} \, Z_0[\eta_w, \bar{\eta}_z],$$
where $Z_0$ is the generating functional for free Dirac fields given by
$$
Z_0[\eta_w, \bar{\eta}_z] = \exp\left(-i\int \bar{\eta}_z \, S(z-w) \, \eta_w \, dz \, dw\right).
$$
Here, $\eta$ and $\bar{\eta}$ are source terms. Also, $S^{-1} = i\gamma\cdot\partial - m$ is the operator appearing in the quadratic term of the lagrangian.
Notation
$\eta_x \equiv \eta(x)$ etc.
At first, I determine $\frac{\delta Z_0}{\delta\bar{\eta}_y}$ as
$$
\frac{\delta Z_0}{\delta\bar{\eta}_y} = -iZ_0 \int S(y-w) \eta_w \, dw. \label{a}\tag{1}
$$
Then I try to compute
\begin{align}
-\frac{\delta^2 Z_0}{\delta\eta_x\delta\bar{\eta}_y} &= -\frac{\delta}{\delta\eta_x} \left[-iZ_0 \int S(y-w) \eta_w \, dw \right] \\
&= i\frac{\delta}{\delta\eta_x} \left[Z_0[\eta_w, \bar{\eta}_z] \int S(y-w) \eta_w \, dw \right] \label{b}\tag{2}
\end{align}
Question 


*

*How to proceed from this step (eq. (\ref{b}))? I have to take the functional derivative of the product of two Grassmann functionals. What is the relevant formula for it? If you also mention any reference, that would be great.

*In eq. (\ref{a}) I have written $Z_0$ before the functional derivative part. Should I write it after the functional derivative term? In other words, what is the chain rule for Grassmann functionals? 

*In the Appendix, I have mentioned the formula to take the functional derivative of a product of Grassmann functionals. Let's say, I have a product of some Grassmann functionals and an ordinary function $f(x) \in \mathbb{C} \forall x$. Then how to evaluate this functional derivative? That is,
$$
\frac{\delta}{\delta\psi(x)} [\psi(y_1) f(y_2) \psi(y_3)] = ?
$$
where $\psi$ is a Grassmann field.
Appendix 
The formula used to compute the eq. (\ref{a}) is given below.
$$
\frac{\delta}{\delta\psi(x)} [\psi(y_1) \cdots \psi(y_n)] = \delta(y_1-x) \psi(y_2) \cdots \psi(y_n) + (-1) \delta(y_2-x) \psi(y_1) \cdots \psi(y_n) + \cdots \cdots + (-1)^{n-1} \delta(y_n-x) \psi(y_1) \cdots \psi(y_{n-1}).
$$
 A: First of all, the functional integral $Z_0$ is a real number, since it is defined as an vacuum expectation value:
$$Z_0[\zeta,\bar{\zeta}]:=\langle0|T\,e^{i\langle\bar{\zeta}_x\psi_x+\bar{\psi}_x\zeta_x\rangle}|0\rangle$$
where $T$ is the time ordering operato and:
$$\langle\bar{\zeta}_x\psi_x+\bar{\psi}_x\zeta_x\rangle:=\int d^4x(\bar{\zeta}(x)\psi(x)+\bar{\psi}(x)\zeta(x))$$
Now, after some analytical steps, it is found that this object must satisfy the Symanzik equation:
$$\left[(i\gamma^\mu\partial_\mu-m)\frac{\delta}{i\delta\bar{\zeta}_z}-\zeta_z\right]Z_0[\zeta,\bar{\zeta}]=0$$
It is found that a solution is readly obtained by putting (as an ansatz):
$$Z_0[\zeta,\bar{\zeta}]=e^{-\int d^4x\int d^4y\,(\bar{\zeta}(x)S_F(x-y)\zeta(y))}=e^{-\langle\bar{\zeta}_xS^F_{xy}\zeta_y\rangle}$$
But, in general, a solution for a linear differential equation can be searched by means of a fourier transform. In this way we define the functional Fourier transform of $Z_0$ as:
$$Z_0[\zeta,\bar{\zeta}]=\int\mathscr{D}\psi\int\mathscr{D}\bar{\psi}\,\tilde{Z}[\psi,\bar{\psi}]e^{i\int d^4x(\bar{\zeta}(x)\psi(x)+\bar{\psi}(x)\zeta(x))}=\int\mathscr{D}\psi\int\mathscr{D}\bar{\psi}\,\tilde{Z}[\psi,\bar{\psi}]e^{i\langle\bar{\zeta}_x\psi_x+\bar{\psi}_x\zeta_x\rangle}$$
By putting this functional fourier transform in the Symanzik equation, we can identify:
$$\tilde{Z}[\psi,\bar{\psi}]:=\mathcal{N}e^{i\int d^4x\,\bar{\psi}(i\gamma^\mu\partial_\mu-m)\psi}=\mathcal{N}e^{iS_D[\psi,\bar{\psi}]}$$
where $\mathcal{N}$ is a constant, and obtain:
$$Z_0[\zeta,\bar{\zeta}]=\mathcal{N}\int\mathscr{D}\psi\int\mathscr{D}\bar{\psi}\,e^{iS_D+i\langle\bar{\zeta}_x\psi_x+\bar{\psi}_x\zeta_x\rangle}$$
Now (in analogy with the bosonic case), the 2n-point Green Function can be written as:
$$S^{(2n)}_0(x_1,...,x_n;y_1,...,y_n)=\langle 0|\psi(x_1)\cdots\psi(x_n)\bar{\psi}(y_1)\cdots\bar{\psi}(y_n)|0\rangle=\frac{\delta^{(2n)}Z_0[\zeta,\bar{\zeta}]}{\delta\bar{\zeta}(x_1)\cdots\delta\bar{\zeta}(x_n)\delta\zeta(y_1)\cdots\delta\zeta(y_n)}\bigg|_{\zeta=0,\bar{\zeta}=0}$$
And if we want to evaluate the 2-point Green function, we first evaluate:
$$\frac{\delta Z_0}{\delta\zeta(x_2)}=\frac{\delta}{\delta\zeta(x_2)}e^{-\int d^4x\int d^4y (\bar{\zeta}(x)S_F(x-y)\zeta(y)}=-\int d^4x\, (-1)\bar{\zeta}(x)S_F(x-x_2)\cdot Z_0[\zeta,\bar{\zeta}]$$
where the $(-1)$ is due to the fact the grassman functional derivative has to jump over the $\bar{\zeta}$, which is a grassman valued function. Then:
$$\frac{\delta}{\delta\bar{\zeta}(x_1)}\left(\frac{\delta Z_0}{\delta\zeta(x_2)}\right)=\frac{\delta}{\delta\bar{\zeta}(x_1)}\left(\int d^4x\, \bar{\zeta}(x)S_F(x-x_2)\cdot Z_0[\zeta,\bar{\zeta}]\right)=S_F(x_1-x_2)\cdot Z_0+\left(\int d^4x\, \bar{\zeta}(x)S_F(x-x_2)\cdot (-1)\int d^4 y\, S_F(x_1-y)\zeta(y)\cdot Z_0\right)$$
By putting $\zeta=0,\bar{\zeta}=0$, the $Z_0$ is $1$ and the second term goes to zero, obtaining:
$$S^{(2)}_0(x_1,x_2)=S_F(x_1-x_2)$$
The functional integral is a real Number, it is not Grassman valued, so you don't have to worry about the order of your equation (2).
A: 
I am trying to compute the 2-point Green function $\tau_2(x,y) = -\frac{\delta^2}{\delta\eta_x \delta \bar{\eta}_y} \, Z_0[\eta, \bar{\eta}]$

You have to take the $\eta=0$/$\bar{\eta}=0 $ limit as the final step 
$$ \tau_2(x,y) = -\frac{\delta^2}{\delta\eta_x \delta \bar{\eta}_y} \, Z_0[\eta, \bar{\eta}] \bigg|_{\eta=0,\bar{\eta}=0}.$$
Green functions (as opposed to the functional integral $Z_0[\eta, \bar{\eta}]$) should not be explicitly  $\eta$/$\bar{\eta}$ dependent. 

  
*
  
*How to proceed from this step (eq. (2)? 
  

The $\frac{\delta}{\delta\eta_x} \left[Z_0[\eta, \bar{\eta}] \right]$ part in eq.(2) will drop out after taking the $\eta=0$/$\bar{\eta}=0$ limit at the final step. Thus you only care about the $\frac{\delta}{\delta\eta_x} \left[\int S(y-w) \eta_w \, dw \right]$ part.


  
*In eq. (1) I have written $Z_0$ before the functional derivative part. Should I write it after the functional derivative term?
  

The key point is that the functional integral $Z_0[\eta, \bar{\eta}]
$ is EVEN-graded in Grassmann numbers $\eta$/$\bar{\eta}$. The order of $Z_0$ in eq. (1) does NOT matter. 


  
*Then how to evaluate this functional derivative?
  

Just treat the ordinary function $f(x)$ as a constant real/complex number. It does not interfere with the functional derivative. 
