Assuming everything is smooth:
$\textbf{Lemma}.$ Let $M$ be a smooth manifold. The map $D: C^{\infty}(M) \to C^{\infty}(M)$ is a derivation if and only if, $Df = Xf$ for some smooth vector field $X$.
In particular, this says that derivations of $C^{\infty}(M)$ can be identified with smooth vector fields. So, it suffices to show that the Lie bracket is a derivation of $C^{\infty}(M)$.
$\textbf{Sketch.}$
$$
\begin{align*}
[X,Y](fg) &= X(Y(fg) - Y(X(fg)) \\
&= X(fYg + gYf) - Y(fXg + gXf) \\
&= \ldots
\end{align*}
$$
The functions on $T^{*}Q$ that are of the form $f_{X}$ for some $X\in\mathfrak{X}(Q)$ are exactly the fiberwise linear functions $C^{\infty}_{lin}(T^{*}Q)$ on $T^{*}Q$. So the $p_{i}$ are of this form, but the $q^{j}$ are not (as they are fiberwise constant).
a) We have that $p_{i}=f_{\partial_{q^{i}}}$, since
$$
p_{i}(dq^{j})=\delta_{i,j}=f_{\partial_{q^{i}}}(dq^{j}).
$$
Hence,
$$\label{1}\tag{1}
\{p_{i},p_{j}\}=\{f_{\partial_{q^{i}}},f_{\partial_{q^{j}}}\}=-f_{[\partial_{q^{i}},\partial_{q^{j}}]}=-f_{0}=0.
$$
b) As said, $q^{j}$ is not fiberwise linear, but $q^{j}p_{i}$ is. Indeed, we have $q^{j}p_{i}=f_{q^{j}\partial_{q^{i}}}$. So
$$\tag{2}\label{2}
\{q^{j}p_{i},p_{k}\}=\{f_{q^{j}\partial_{q^{i}}},f_{\partial_{q^{k}}}\}=-f_{[q^{j}\partial_{q^{i}},\partial_{q^{k}}]}=f_{\frac{\partial q^{j}}{\partial q^{k}}\partial_{q^{i}}}=f_{\delta_{j,k}\partial_{q^{i}}}=\delta_{j,k}f_{\partial_{q^{i}}}=\delta_{j,k}p_{i}.
$$
On the other hand, using the Leibniz rule of the Poisson bracket $\{\cdot,\cdot\}$, we have
$$\tag{3}\label{3}
\{q^{j}p_{i},p_{k}\}=q^{j}\{p_{i},p_{j}\}+p_{i}\{q^{j},p_{k}\}=p_{i}\{q^{j},p_{k}\},
$$
using \eqref{1} in the last equality. Comparing \eqref{2} and \eqref{3} then gives
$$\tag{4}\label{4}
\{q^{j},p_{k}\}=\delta_{j,k}.
$$
c) At last, for the fiberwise linear vector fields $q^{k}p_{i}=f_{q^{k}\partial_{q^{i}}}$ and $q^{l}p_{j}=f_{q^{l}\partial_{q^{j}}}$, we get
\begin{align}
\{q^{k}p_{i},q^{l}p_{j}\}&=\{f_{q^{k}\partial_{q^{i}}},f_{q^{l}\partial_{q^{j}}}\}=-f_{[q^{k}\partial_{q^{i}},q^{l}\partial_{q^{j}}]}=-f_{q^{k}\frac{\partial q^{l}}{\partial q^{i}}\partial_{q^{j}}-q^{l}\frac{\partial q^{k}}{\partial q^{j}}\partial_{q^{i}}}
=-\delta_{l,i} q^{k} f_{\partial_{q^{j}}}+\delta_{k,j}q^{l}f_{\partial_{{q}^{i}}}\\
&=-\delta_{l,i} q^{k} p_{j}+\delta_{k,j}q^{l}p_{i}.\tag{5}\label{5}
\end{align}
But also, by the Leibniz rule:
\begin{align}
\{q^{k}p_{i},q^{l}p_{j}\}&=\{q^{k},q^{l}\}p_{i}p_{j}+\{q^{k},p_{j}\}p_{i}q^{l}+\{p_{i},q^{l}\}q^{k}p_{j}+\{p_{i},p_{j}\}q^{k}q^{l}\\
&=\{q^{k},q^{l}\}p_{i}p_{j}+\delta_{k,j}p_{i}q^{l}-\delta_{i,l}q^{k}p_{j},\tag{6}\label{6}
\end{align}
using \eqref{1} and \eqref{4} in the last equality. Comparing \eqref{5} and \eqref{6} then gives
$$\tag{7}\label{7}
\{q^{k},q^{l}\}=0.
$$
The equalities \eqref{1},\eqref{4} and \eqref{7} now show that $\{\cdot,\cdot\}$ is indeed the canonical Poisson bracket.
Best Answer
The coefficients $a_{12},a_{13}, a_{23}\in C^{\infty}(\mathbb{R}^3)$ need to satisfy the following equation \begin{align*}\sum_{l=1}^3 a_{lk}\partial_la_{ij}+a_{li}\partial_l a_{jk} +a_{lj}\partial_la_{ki} =0 \end{align*} for all $i,j,k\in \{1,2,3\}$. As mentinoed in the comments it is induced by evalutating the Jacobi identity on the coordinate functions $x_i$. Here $a_{ji}:=-a_{ij}$ for $i\le j$.