How can I derive a commutator between derivative of a scalar field and Lorentz generator $M$? Let's suppose we  have
$$
U(\Lambda)^{-1}\partial^\mu\phi(x) U(\Lambda)=\Lambda^\mu_{\;\rho}\bar\partial^\rho\phi(\Lambda^{-1}x)
$$
where $\bar\partial^\rho$ is derivative with respect to the argument $\Lambda^{-1}x$, $U(\Lambda)$ is the unitary operator of Lorentz group and $\phi(x)$ is a scalar field. 
Using infinitesimal transformation $\Lambda = 1 + \delta\omega$ and knowing
$$
U(\Lambda)= U(1+\delta\omega) = I+\frac{1}{2\hbar}\delta\omega_{\mu\nu}M^{\mu\nu}\\\Lambda^\rho_{\;\tau}= \delta^\rho_{\;\tau}+\frac{i}{2\hbar}\delta\omega_{\mu\nu}\big(S^{\mu\nu}_V\big)^\rho_{\;\tau}
$$
where V stands for vector representation, how can I derive this following commutator?
$$
\big[\partial^\rho\phi(x), M^{\mu\nu}\big]= L^{\mu\nu}\partial^\rho\phi(x)+\big(S^{\mu\nu}_V\big)^\rho_{\;\tau}\partial^\tau\phi(x)
$$
with
$$
L^{\mu\nu}=\frac{\hbar}{i}\big(x^\mu\partial^\nu-x^\nu\partial^\mu\big)\\ \big(S^{\mu\nu}_V\big)^\rho_{\;\tau}=\frac{\hbar}{i}\big(g^{\mu\rho}\delta^\nu_{\;\tau}-g^{\nu\rho}\delta^\mu_{\;\tau}\big)
$$
In particular, how can I translate $\bar\partial$ in $\partial$ in the commutator?
 A: I believe that you're working with the second section of Srednicki so I will try to be close to the notation of that book. (Hence I believe that your expression $U(\Lambda)= U(1+\delta\omega) = I+\frac{1}{2\hbar}\delta\omega_{\mu\nu}M^{\mu\nu}$ should have been $I+\frac{i}{2\hbar}\delta\omega_{\mu\nu}M^{\mu\nu}$)
First, note that $U(\Lambda)^{-1}\phi(x)U(\Lambda)=\phi(\Lambda^{-1}x)$. Take $\Lambda^\mu_{\; \nu}=\delta^\mu_{\; \nu}+\delta\omega^\mu_{\; \nu}$ and $U(\Lambda)=I+\frac{i}{2\hbar}\delta\omega_{\mu\nu}M^{\mu\nu}$. The left hand side is just $\phi(x)+\frac{i}{2\hbar}\delta \omega_{\mu\nu}[\phi(x),M^{\mu\nu}]$ up to linear order of $\delta\omega_{\mu\nu}$. The right hand side is $\phi(x^\mu+\delta\omega^\mu_{\;\nu}x^\nu)=\phi(x)+\delta\omega^\mu_{\;\nu}x^\nu\partial_\mu\phi(x)=\phi(x)+\delta\omega_{\mu\nu}x^\nu\partial^\mu\phi(x)$. Compare the coefficients $\delta\omega_{\mu\nu}$ and use the antisymmetry, we have $[\phi(x),M^{\mu\nu}]=\frac{\hbar}{i}(x^\mu\partial^\nu-x^\nu\partial^\mu)$.
Now $\partial^\rho(U(\Lambda)^{-1}\phi(x)U(\Lambda))=U(\Lambda)^{-1}\partial^\rho \phi(x)U(\Lambda)=\partial^\rho\phi(\Lambda^{-1}x)$ which is equivalent to your first equation.
Again, use $U(\Lambda)= I+\frac{i}{2\hbar}\delta\omega_{\mu\nu}M^{\mu\nu}$, the equation above is $[\partial^\rho\phi(x),M^{\mu\nu}]=\partial^\rho(\frac{\hbar}{i}(x^\mu\partial^\nu-x^\nu\partial^\mu)) \phi(x)$. Following yours (and Srednicki's) notation, $L^{\mu\nu}=\frac{\hbar}{i}(x^\mu\partial^\nu-x^\nu\partial^\mu)$. Note that $[\partial^\rho\phi(x),M^{\mu\nu}]=\partial^\rho L^{\mu\nu}\phi(x)=L^{\mu\nu}\partial^{\rho}\phi(x)+[\partial^\rho,L^{\mu\nu}]\phi(x)$; hence your equation $\big[\partial^\rho\phi(x), M^{\mu\nu}\big]= L^{\mu\nu}\partial^\rho\phi(x)+\big(S^{\mu\nu}_V\big)^\rho_{\;\tau}\partial^\tau\phi(x)$ is equivalent to proving that $[\partial^\rho,L^{\mu\nu}]=\big(S^{\mu\nu}_V\big)^\rho_{\;\tau}\partial^\tau$.
But \begin{aligned}\frac{\hbar}{i}[\partial^{\rho},x^\mu\partial^\nu-x^\nu\partial^\mu]&=\frac{\hbar}{i}([\partial^\rho,x^\mu]\partial^\nu-[\partial^\rho,x^\nu]\partial^\mu)=\frac{\hbar}{i}(g^{\rho\mu}\partial^\nu-g^{\rho\nu}\partial^\mu)\\&=\frac{\hbar}{i}(g^{\rho\mu}\delta^\nu_{\;\tau}-g^{\rho\nu}\delta^\mu_{\;\tau})\partial^\tau=\big(S^{\mu\nu}_V\big)^\rho_{\;\tau}\partial^\tau\end{aligned} by your definition.
