Fermion fields anticommute because they are Grassmann numbers, that is, \begin{equation} \psi \chi = - \chi \psi. \end{equation} I was wondering whether derivatives with respect to Grassmann numbers also anticommute, as in
\begin{equation} \frac{\partial}{\partial \psi} (\bar \psi \psi) \stackrel{?}{=} - \bar \psi \frac{\partial}{\partial \psi} (\psi) = - \bar \psi. \end{equation}
Since$^1$ $$(\frac{\partial}{\partial z}z)~=~1\tag{1}$$ for any supernumber-valued variable $z$, the Grassmann-parity of the partial derivative $\frac{\partial}{\partial z}$ should be the same as the Grassmann-parity of $z$ in order for eq. (1) to preserve Grassmann-parity.
In superspace $\mathbb{R}^{n|m}\ni(x,\theta)$ a functional derivative $\frac{\delta}{\delta z(x,\theta)}$ and its superfield $z(x,\theta)$ carry the same (opposite) Grassmann parity if the number $m$ of $\theta$'s is even (odd), respectively: $$ (\frac{\delta}{\delta z(x,\theta)}z(x^{\prime},\theta^{\prime}))~=~\delta^n(x\!-\!x^{\prime})\delta^m(\theta\!-\!\theta^{\prime}) \tag{2}.$$
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$^1$ The parenthesis on the left-hand side of eq. (1) is supposed to indicate that the derivative $\frac{\partial}{\partial z}$ does not act past $z$.
In a Grassmann algebra (or more pedantic, a $\mathbb{Z}_2$-graded algebra), the expression with two equality signs you wrote is ill-defined. One either has a left-derivative, or a right-derivative, which are different operators in terms of results. More precisely,
$$ \frac{\partial^L}{\partial \psi} \left(\bar{\psi}\psi\right) = (-)^{\epsilon(\bar{\psi})\epsilon(\psi)} \bar{\psi} $$
$$ \frac{\partial^R}{\partial \psi} \left(\bar{\psi}\psi\right) = \bar{\psi} $$,
where the epsilons are the Grassmann parities of the two variables. if both variables are Grassmann-odd, then the results of the operators are different (one is minus the other).
