= Solution
Write
$$
J^\alpha{}_{\mu}=\frac{\partial\widetilde x^\alpha}{\partial x^\mu}.
$$
By the <chain rule>, $\widetilde V^\alpha=J^\alpha{}_{\mu}V^\mu$, $\widetilde W^\alpha=J^\alpha{}_{\mu}W^\mu$, and $\widetilde V^\beta\widetilde\partial_\beta=V^\nu\partial_\nu$. Therefore
$$
\begin{aligned}
[\widetilde V,\widetilde W]^\alpha
&=V^\nu\partial_\nu(J^\alpha{}_{\mu}W^\mu)-W^\nu\partial_\nu(J^\alpha{}_{\mu}V^\mu)\\
&=J^\alpha{}_{\mu}[V,W]^\mu+(V^\nu W^\mu-W^\nu V^\mu)\frac{\partial^2\widetilde x^\alpha}{\partial x^\nu\partial x^\mu}.
\end{aligned}
$$
The last term vanishes because its first factor is antisymmetric in $\mu,\nu$ and its second is symmetric. Thus the <coordinate invariance of the Lie bracket> is
$$
\boxed{[\widetilde V,\widetilde W]^\alpha=J^\alpha{}_{\mu}[V,W]^\mu.}
$$
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