= Solution
On <smooth functions>, the <Lie derivative of a function> satisfies
$$
[\mathcal L_X,\mathcal L_Y]f=X(Y(f))-Y(X(f))=[X,Y](f)=\mathcal L_{[X,Y]}f.
$$
On an arbitrary <vector field> $W$, the <Lie derivative of a vector field> gives
$$
[\mathcal L_X,\mathcal L_Y]W=[X,[Y,W]]-[Y,[X,W]]=[[X,Y],W]=\mathcal L_{[X,Y]}W.
$$
The middle equality follows from the <Jacobi identity> for the <Lie bracket of vector fields>. One can verify that identity without assuming the result being proved: view <vector fields> as derivations acting on <smooth functions>, expand their <commutators>, and cancel the six compositions. Equality as derivations implies equality as <vector fields>.
For the second identity, let $A,B,C$ be any operators on either <smooth functions> or <vector fields>. Associativity of composition gives
$$
\begin{aligned}
[[A,B],C]&=ABC-BAC-CAB+CBA,\\
[[B,C],A]&=BCA-CBA-ABC+ACB,\\
[[C,A],B]&=CAB-ACB-BCA+BAC.
\end{aligned}
$$
Each term cancels in the sum. Applying this to $A=\mathcal L_X$, $B=\mathcal L_Y$, $C=\mathcal L_Z$ proves \b[both requested identities on both kinds of arguments]:
$$
\boxed{[\mathcal L_X,\mathcal L_Y]=\mathcal L_{[X,Y]},\qquad
[[\mathcal L_X,\mathcal L_Y],\mathcal L_Z]+[[\mathcal L_Y,\mathcal L_Z],\mathcal L_X]+[[\mathcal L_Z,\mathcal L_X],\mathcal L_Y]=0.}
$$
The <commutator identity for Lie derivatives> expresses that the <Lie derivative of a tensor field> represents the <Lie bracket of vector fields>.
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