Understanding the Functional Poisson Bracket In classical field theory (for a single field $\psi$) the dynamical variables are defined to be functions of the fields $\psi$, $\pi$, $\partial_{x_{i}}\psi$ and maybe $\mathbf{r}$, where $\pi$ is the conjugated field to $\psi$.
For $F=\int\mathcal{F}\,d\mathbf{r}$ and $G=\int\mathcal{G}\,d\mathbf{r}$, where F and G are dynamical variables, the functional Poisson bracket can be defined according to (José and Saletan, “Classical Dynamics: A Contemporary Approach”, cap 9))
$$\left\{ F,G\right\} ^{f}=\intop\left(\frac{\delta F}{\delta\psi}\frac{\delta G}{\delta\pi}-\frac{\delta F}{\delta\pi}\frac{\delta G}{\delta\psi}\right)d\mathbf{x},$$ where the derivatives are functional derivatives. The fields themselves have the canonical property
$$\left\{ \psi(\mathbf{y}),\pi(\mathbf{z})\right\}  =\delta(\mathbf{y}-\mathbf{z}),$$
$$\left\{ \pi(\mathbf{y}),\pi(\mathbf{z})\right\}  =\left\{ \psi(\mathbf{y}),\psi(\mathbf{z})\right\} =0.$$
So far so good, but I'm not sure how to handle the functional derivatives. I'm interested, for example, in the following bracket
$$\left\{ F(\mathbf{x}),\pi(\mathbf{z})\right\} ^{f}$$ $(F(\mathbf{x})\equiv F(\psi(\mathbf{x}),\pi(\mathbf{x}),\partial_{x_{i}}\psi(\mathbf{x}))$
Using $\frac{\delta\pi(\mathbf{z})}{\delta\psi}=0$ and $\frac{\delta\pi(\mathbf{z})}{\delta\pi}=\delta(\mathbf{y}-\mathbf{x})$, I think the answer is
$$\left\{ F(\mathbf{x}),\pi(\mathbf{z})\right\} ^{f}=\frac{\delta F(\mathbf{z})}{\delta\psi}.$$
Is this result correct?
 A: You are on the right track, but some clarification is needed. First as you mentioned, the Poisson bracket is defined over functionals of the local fields, i.e. an object of the form $F=\int d^n x {\cal F}[\psi(x),\pi(x)]$. Note that here we integrate over $x$, so $F$ is not a function of $x$. Accordingly, your last equation should read
\begin{align}
\{F,\pi(z)\}&=\int d^nx \{{\cal F}[\psi(x),\pi(x)],\pi(z)\}=\int d^nx \frac{\delta {\cal F}}{\delta \psi(x)}\frac{\delta \pi(z)}{\delta \pi(x)}=\int d^nx \frac{\delta {\cal F}}{\delta \psi(x)}\delta^n(z-x)=\frac{\delta {\cal F}[\psi(z),\pi(z)]}{\delta \psi(z)}
\end{align}
So in your last equation, there is two typos: on the lhs, there is no argument $x$ for $F$, and on the rhs, it is the density $\cal{F}$ instead of the functional $F$.
If you insist on having the left hand side of your last equation as it is, you can start by smearing the density $\cal{F}$ with a delta function, i.e. you take your functional to be $F(x)=\int d^nx' \delta^n(x'-x){\cal F}[\psi(x'),\pi(x')]$. Now repeating the procedure you find
\begin{align}
\{F(x),\pi(z)\}&=\frac{\delta {\cal F}[\psi(z),\pi(z)]}{\delta \psi(z)}\,\delta^n(x-z).
\end{align}
