proof of the Jacobi Identity for certain poisson brackets I have to prove that these are effectively Poisson bracket. Specifically that the satisfy Jacobi Identity when $a_{ij}=-a_{ji}$. $$ \left\{ f,g\right\} =\stackrel{\scriptscriptstyle i,j=1..3}{\sum}\left(a_{ij}+\stackrel{\scriptscriptstyle k=1..3}{\sum}\epsilon_{ijk}x^{k}\right)\frac{\partial f}{\partial x^{i}}\frac{\partial g}{\partial x^{j}}, $$
I tried the plain and direct way but it involves pages of calculus... so I thought: maybe is there a smart way that I didn't see to prove it?

Yes, it is much easier than doing pages of calculus:
In coordinates it always suffices to show the Jacobi identity for coordinate functions $(f,g,h)=(x_i, x_j, x_k)$ with $i<j<k$. Since we are on $\mathbb{R}^3$ we only have to show it for $(x_1,x_2,x_3)$: $$\{\{x_1,x_2\},x_3\} + \{\{x_2,x_3\},x_1\} + \{\{x_3,x_1\},x_2\}=0.$$ For the "inner" brackets you can leave out the constants $a_{ij}$ because they will be differentiated away by the "outer" brackets. Then in each term the "inner" bracket will be (up to irrelevant constants) equal to $\pm$ the other argument in the "outer" bracket (the sum over $k$ contains only one nonzero term), so each term is zero by skew-symmetry.
By the way this is the sum of two Poisson structures, a constant one and the usual one on $\mathfrak{so}(3)^*$ with Casimir $\tfrac{1}{2}(x_1^2 + x_2^2 + x_3^2)$; the Jacobi identity for the sum is equivalent to their compatibility.
Since you tagged representation theory: the constant bracket defines a linear map $C: \mathfrak{so}(3) \wedge \mathfrak{so}(3) \to \mathbb{R}$ by $(x_i,x_j) \mapsto a_{ij}$ and compatibility is equivalent to saying that $C$ is a $2$-cocycle in the cohomology of $\mathfrak{so}(3)$ associated with the trivial representation of $\mathfrak{so}(3)$ on $\mathbb{R}$.
