Problem 6 of Sheet 1 - Quantum field theory David Tong - Variation of Lagrangian density The problem reads:

Consider the infinitesimal form of the Lorentz transformation derived in the previous question: $x^\mu \rightarrow x^\mu +\omega^{\mu}_\nu x^\nu$. Show that the scalar field transforms as $$\phi(x)\rightarrow \phi'(x)=\phi(x)-\omega^{\mu}_\nu x^\nu \partial_\mu \phi(x)$$
and hence show that the variation of the Lagrangian density is a total derivative $$\delta \mathcal{L}=-\partial_\mu (\omega^{\mu}_\nu x^\nu \mathcal{L})$$

We consider the transformation as
$$x^\mu \rightarrow x'^\mu=x^\mu +\delta x^\mu \\
\phi\rightarrow \phi'=\phi+\delta \phi \\
\partial_\mu \phi\rightarrow (\partial_\mu\phi)'=(\partial_\mu \phi)+\delta (\partial_\mu \phi)$$
Using $$\delta \mathcal{L}=\mathcal{L}\left(\phi(x_\mu),\partial_\mu (\phi_\mu),x_\mu\right)-\mathcal{L}\left(\phi'(x'_\mu),\partial_\mu \phi'_\mu(x'_\mu),x'_\mu\right)$$
Using the Taylor expansion, we find that ( See No-Nonsense QFT Eq. 4.32 )
$$\delta \mathcal{L}=-\partial_\mu \left(\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)}\delta \phi\right)-\frac{\partial \mathcal{L}}{\partial x_\mu}\delta x_\mu $$
Putting the values, we have
$$\delta \mathcal{L}=-\partial_\mu \left(\frac{\partial \mathcal{L}}{\partial(\partial_\mu \phi)} \omega^\lambda_\nu x^\nu \partial_\lambda \phi\right)-\frac{\partial \mathcal{L}}{\partial x_\mu} \omega^\mu_\nu x^\nu$$
I don't find any further simplification. Can you help me on how to proceed?
 A: You can use the fact that $\mathcal{L}$, is a scalar field.  Thus $\delta \mathcal{L}$ is of the same form as $\delta \phi$ just by replacing $\phi \rightarrow \mathcal{L}$. We further use the fact that $\omega$ is antisymmetric. If you use these two hints, it should be enough to solve the problem.
The rest of the answer is about how the calculation actually proceeds, if you need it. I do however recommend trying it first using just the small hints above:
With this, you get $\delta \mathcal{L} = - {\omega^\mu}_\nu x^\nu\partial_\mu \mathcal{L}$. Now we note that $\omega$ is antisymmetric, i.e. $\omega_{\mu\nu} = - \omega_{\nu\mu} \implies Tr(\omega) = 0$. Then $\delta \mathcal{L} = - {\omega^\mu}_\nu x^\nu\partial_\mu \mathcal{L} = - \partial_\mu ({\omega^\mu}_\nu x^\nu \mathcal{L}) + \delta^\nu_\mu \omega^\mu_\nu \mathcal{L}$. Now using the fact that the trace of $\omega$ vanishes, you get the desired result.
A: I will try to complement the answer by Thomas, since some things were not that clear to me.
We start by varying the Lagrangian $\mathcal{L}(\phi(x),\partial_{\mu}\phi(x),x^{\mu}):$
$$ \begin{split}\delta \mathcal{L} &= \frac{\partial \mathcal{L}}{\partial \phi}\delta \phi + \frac{\partial \mathcal{L}}{\partial (\partial_{\mu}\phi)}\delta(\partial_{\mu}\phi)+\frac{\partial \mathcal{L}}{\partial x^{\mu}}\delta x^{\mu} \\ &=\partial_{\mu} \bigg{(}\frac{\partial \mathcal{L}}{\partial(\partial_{\mu}\phi)}\delta \phi\bigg{)}+ \frac{\partial \mathcal{L}}{\partial x^{\mu}}\delta x^{\mu},\end{split}\tag{1}$$
where in the second line we used the chain rule and imposed the equation of motion.  Then, the first term vanishes, because the Lagrangian should be a Lorentz scalar.  Generally, the Lagrangian is a Lorentz scalar if and only if $$\delta \mathcal{L}=\delta x^{\mu}\partial_{\mu}\mathcal{L}.\tag{2}$$
So in the end, we just have
$$\delta \mathcal{L}=\frac{\partial \mathcal{L}}{\partial x^{\mu}}\delta x^{\mu}=-{\omega^{\ \mu}}_{\nu}x^{\nu}\partial_{\mu} \mathcal{L},\tag{3}$$
where the minus comes from doing an active rather than a passive transformation.
The result then follows trivially using the property that ${\omega^{\ \mu}}_{\mu}=0$.
Many thanks to ACuriousMind who helped me with this in The h Bar.
A: In the  last equation in your question, the term involving  $\partial {\mathcal L}/{\partial x^\mu}$ is zero becuase  ${\mathcal L}$ does not depend explicitly on $x^\mu$. Assuming that the other term is correct, you are done.
