I'm working through Introduction to Lie Algebras by Erdmann and Wildon. In Chapter 3, Exercise 3.9 asks the reader to develop the semidirect product of Lie algebras. I got very stuck on the verification of the Jacobi identity, but the link here suggested the hint/trick of breaking the identity down so there are zeros in each pair, which gets you the required verification. However...
Recalling that the Lie bracket is defined as: $$ [(s_1,x_1),(s_2,x_2)]=([s_1,s_2],[x_1,x_2]+\theta(s_1)(x_2)-\theta(s_2)(x_1)), $$ where $\theta:S\rightarrow \text{Der}(X)$, I did the obvious thing and try to push the identity through as far as I could, simplifying along the way, and the most simplification I could achieve was to get: $$ (0,\theta(s_1)[x_2,x_3]+\theta(s_2)[x_3,x_1]+\theta(s_3)[x_1,x_2]), $$ which, if the Jacobi holds, must equal 0, which implies that in any semidirect product, this is true: $$ \theta(s_1)[x_2,x_3]+\theta(s_2)[x_3,x_1]+\theta(s_3)[x_1,x_2]=0. $$
I couldn't figure out how to prove this, but is this an actual known identity in semidirect products? That seems like a very restrictive condition.
For reference, here are the steps of my calculation. Start here:
$$ [(s_1,x_1),[(s_2,x_2),(s_3,x_3)]]+[(s_2,x_2),[(s_3,x_3),(s_1,x_1)]]+[(s_3,x_3),[(s_1,x_1),(s_2,x_2)]]. $$
Then we have: $$ [(s_2,x_2),(s_3,x_3)]=([s_2,s_3],[x_2,x_3]+\theta(s_2)(x_3)-\theta(s_3)(x_2)), $$
$$ [(s_3,x_3),(s_1,x_1)]=([s_3,s_1],[x_3,x_1]+\theta(s_3)(x_1)-\theta(s_1)(x_3)), $$
$$ [(s_1,x_1),(s_2,x_2)]=([s_1,s_2],[x_1,x_2]+\theta(s_1)(x_2)-\theta(s_2)(x_1)). $$
Then,
$$ [(s_1,x_1),([s_2,s_3],[x_2,x_3]+\theta(s_2)(x_3)-\theta(s_3)(x_2))]= ([s_1,[s_2,s_3]],[x_1,[x_2,x_3]]+\theta(s_1)([x_2,x_3]+\theta(s_2)(x_3)-\theta(s_3)(x_2))-\theta([s_2,s_3])(x_1)), $$
$$ [(s_2,x_2),([s_3,s_1],[x_3,x_1]+\theta(s_3)(x_1)-\theta(s_1)(x_3))]= ([s_2,[s_3,s_1]],[x_2,[x_3,x_1]]+\theta(s_2)([x_3,x_1]+\theta(s_3)(x_1)-\theta(s_1)(x_3))-\theta([s_3,s_1])(x_2)), $$
$$ [(s_3,x_3),([s_1,s_2],[x_1,x_2]+\theta(s_1)(x_2)-\theta(s_2)(x_1))]= ([s_3,[s_1,s_2]],[x_3,[x_1,x_2]]+\theta(s_3)([x_1,x_2]+\theta(s_1)(x_2)-\theta(s_2)(x_1))-\theta([s_1,s_2])(x_3)). $$
Things start to go our way, where you recognize the Jacobi identity for the underlying Lie algebras. Adding across components, we get that $[s_1,[s_2,s_3]]+[s_2,[s_3,s_1]]+[s_3,[s_1,s_2]]=0$ and $[x_1,[x_2,x_3]]+[x_2,[x_3,x_1]]+[x_3,[x_1,x_2]]=0$. In the second component, here's what's "left:"
$$ \theta(s_1)([x_2,x_3]) + \theta(s_1)\circ\theta(s_2)(x_3) -\theta(s_1)\circ\theta(s_3)(x_2) - \theta([s_2,s_3])(x_1) + \theta(s_2)([x_3,x_1]) + \theta(s_2)\circ\theta(s_3)(x_1) - \theta(s_2)\circ\theta(s_1)(x_3) - \theta([s_3,s_1])(x_2) + \theta(s_3)([x_1,x_2]) + \theta(s_3)\circ\theta(s_2)(x_2) - \theta(s_3)\circ\theta(s_2)(x_1) - \theta([s_1,s_2])(x_3). $$
Rearranging:
$$ \theta(s_1)([x_2,x_3])+\theta(s_2)([x_3,x_1])+\theta(s_3)([x_1,x_2]) + \theta(s_1)\circ\theta(s_2)(x_3) - \theta(s_2)\circ\theta(s_1)(x_3) + \theta(s_2)\circ\theta(s_3)(x_1) - \theta(s_3)\circ\theta(s_2)(x_1) + \theta(s_3)\circ\theta(s_1)(x_2) - \theta(s_1)\circ\theta(s_3)(x_2) - \theta([s_2,s_3])(x_1) - \theta([s_3,s_1])(x_2) - \theta([s_1,s_2])(x_3). $$
Things cancel, for example:
$$ \theta(s_1)\circ\theta(s_2)(x_3) - \theta(s_2)\circ\theta(s_1)(x_3) - \theta([s_1,s_2])(x_3) = \theta([s_1,s_2])(x_3)-\theta([s_1,s_2])(x_3)=0. $$