= Solution
The <chain rule> yields
$$
\{F\circ f,G\circ f\}_{q,p}
=(\nabla F\circ f)^T AJA^T(\nabla G\circ f).
$$
If $f$ is a <canonical transformation>, the <symplectic matrix> identity $AJA^T=J$ makes this expression $\{F,G\}_{Q,P}\circ f$. Thus preservation of the <symplectic form> implies preservation of every <Poisson bracket>, for all $C^1$ functions.
Conversely, assume this <Poisson bracket> identity for every $F,G$. Take $F(Z)=u\cdot Z$ and $G(Z)=v\cdot Z$, for arbitrary constant <vectors> $u,v\in\mathbb R^{2n}$. It gives $u^TAJA^Tv=u^TJv$. Since this holds for every $u,v$, $AJA^T=J$, hence $A^TJA=J$. Therefore \b[preservation of all <Poisson brackets> is equivalent to canonicity]. Only first <derivatives> of $f,F,G$ enter this argument.
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