Derivation of Grassmann valued functional I'm trying to evaluate
$$\frac{\delta}{\delta \eta(x)}e^{-\int dz \theta^*(z)\eta(z)}$$
Where $\theta^*(x)$ and $\eta(x)$ are Grassmann valued functions. The context of the functional is in term of functions that can be expressed in a basis of orthonormal functions $\{f_n(x)\}$ such that 
$$\eta(x)=\sum_n\eta_n f_n(x)$$ 
where $\eta_n$ is Grassmann number, which means that for any two Grassmann numbers, $a$, $b$ then $ab=-ba$ and so $a^2=0$, etc.
The definition of the derivative is such that
$$\frac{\delta}{\delta \eta(x)}\eta(y)=\delta(x-y)$$
With that it is easy to show that 
$$\frac{\delta}{\delta \eta(x)}e^{-\int dz \theta^*(z)\eta(z)}=\theta^*(x)$$
However, this should be also equal to 
$$\theta^*(x)e^{-\int dz \theta^*(z)\eta(z)}$$
Which should be possible to show by noticing the property that $a^2=0$ (I think), so by a Taylor expansion
$$\theta^*(x)=^?\theta^*(x)-\int \theta^*(x) \theta^*(z)\eta(z)dz=\theta^*(x)e^{-\int dz \theta^*(z)\eta(z)}$$
But I'm failing to see how the integral is zero. I tried working out whether $\theta^*(x)\theta^*(z)$ is zero alone by expanding the functions in the basis:
$$\theta^*(x)\theta^*(z)=\sum_n \theta^2_nf^*_n(x)f^*_n(y)+\sum_{n<m}\theta_n \theta_m (f^*_n(x)f^*_m(y)-f^*_m(x)f^*_n(y))$$
The only properties of the $f$ I know are the orthogonality:
$$\int dx f^*_m(x)f_n(x)=\delta_{mn}$$
and the completeness condition
$$\sum_n f^*_n(x)f_n(y)=\delta(x-y)$$
But I don't think I can use those.
 A: I know this is late, but better late than never. I think the statement 
\begin{equation}\frac{\delta}{\delta \eta (x)}e^{-\int \mathrm{d}z \theta^{*}(z)\eta(z)}=\theta^{*} (x)\end{equation} is wrong. The exponential doesn't disappear when you differentiate, so it should read 
\begin{equation}\frac{\delta}{\delta \eta (x)}e^{-\int \mathrm{d}z \theta^{*}(z)\eta(z)}=e^{-\int \mathrm{d}z \,\theta^{*}(z)\eta(z)}\left(\frac{\delta}{\delta \eta (x)} \int \mathrm{d}z\, \eta(z) \theta^{*}(z)\right)\end{equation}
Here I've switched $\eta(z)$ and $\theta^{*}(z)$ to get rid of the minus sign. (You could also note that Grassmann derivatives also anticommute, so $\frac{\delta}{\delta \eta (x)}\theta^{*}(z)=-\theta^{*}\frac{\delta}{\delta \eta (x)}$ to get rid of the negative sign). 
The next step is simply evaluating the functional derivative, which according to the property 
\begin{equation}\frac{\delta}{\delta \eta (x)} \eta(z)=\delta(x-z)\end{equation}
you've listed just gives a final result of
\begin{equation}\frac{\delta}{\delta \eta (x)}e^{-\int \mathrm{d}z \theta^{*}(z)\eta(z)}=e^{-\int \mathrm{d}z \theta^{*}(z)\eta(z)}\theta^{*}(x)\end{equation}
Another potential source of confusion could be how to deal with a Taylor expansion of 
\begin{equation}e^{-\int \mathrm{d}z\, \eta(z)\theta^{*}(z)\eta (z)}\end{equation}
It goes just like you think it would, but remember when you raise an integral to a power, you have to integrate over a separate region for each power. For example 
\begin{equation}\left(\int \mathrm{d}z\,\theta^{*}(z)\eta(z)\right)^{2}=\int \mathrm{d}z \int \mathrm{d}x\ (\theta^{*}(z)\eta(z)\theta^{*}(x)\eta(x))\end{equation}
Here, $\eta(z)$ and $\eta(x)$ are different Grassmann variables, so that $\eta(z)\eta(x)\neq 0$ and the integral doesn't vanish. 
Hope this helps, even this late. 
