= Solution
Expand $T=T^\mu{}_{\nu}\,\partial_\mu\otimes dx^\nu$. Using the <tensor Lie derivative> of functions, vectors and covectors and its <Leibniz rule> gives
$$
\boxed{(\mathcal L_XT)^\mu{}_{\nu}
=X^\rho\partial_\rho T^\mu{}_{\nu}
-T^\rho{}_{\nu}\partial_\rho X^\mu
+T^\mu{}_{\rho}\partial_\nu X^\rho.}
$$
The contravariant slot has a minus sign and the covariant slot a plus sign.
For the <commutator identity for Lie derivatives>, put $D=[\mathcal L_X,\mathcal L_Y]-\mathcal L_{[X,Y]}$. The commutator of two <tensor derivations> is itself a <tensor derivation>, and $D$ commutes with <tensor contractions>. On a function, $Df=XYf-YXf-[X,Y]f=0$. On a <vector field> $Z$,
$$
DZ=[X,[Y,Z]]-[Y,[X,Z]]-[[X,Y],Z]=0
$$
by the <Jacobi identity>, which follows here by expanding the commutators of the operators $X,Y,Z$ acting on functions. For a type $(1,1)$ tensor, $T(Z)$ is a <vector field>, and contraction compatibility gives
$$
0=D(T(Z))=(DT)(Z)+T(DZ)=(DT)(Z).
$$
As this holds for every $Z$, $DT=0$. Therefore
$$
\boxed{\mathcal L_X\mathcal L_YT-\mathcal L_Y\mathcal L_XT=\mathcal L_{[X,Y]}T.}
$$
The derivation argument also establishes the identity for arbitrary tensor types by applying it to covector–vector pairings and then to <tensor products>.
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