Variational derivatives of strongly connected diagrams functional in gauge theory Background
In Jorge C. Romao's "Advanced Quantum Field Theory", at the end of page 218, Eq (6.266) reads:
$$\tag{1}
\left.\frac{\delta^{2}}{\delta \omega^{b}(y)\delta A_{\mu}^{c}(z)}\left[ \frac{\delta \Gamma}{\delta K_{\mu}^{a}(x)}\frac{\delta \Gamma}{\delta A^{\mu , a}(x)} \right]\right|_{\text{sources} = 0} = \left.\frac{\delta^{2} \Gamma}{\delta \omega^{b}(y)\delta K_{\mu}^{a}(x)}\frac{\delta^{2} \Gamma}{\delta A^{c}_{\nu}(z)\delta A^{\mu,a}(x)}\right|_{\text{sources} = 0}.
$$
Here
$$\tag{2}
S = \int \mathrm{d}^{4}x\left( -\frac{1}{4}F_{\mu \nu}^{a}F^{\mu \nu}_{a} - B_{a}f^{a} + \frac{1}{2 \varepsilon}B_{a}B^{a} - \bar{\omega}^{a}M_{ab}\omega^{b} + K_{\mu}^{a}\delta A_{a}^{\mu} + L_{a}\delta \omega^{a} \right),
$$
where:


*

*$B_{a}$ is support scalar field which is needed for BRST transformations

*$f_{a}$ is the gauge condition for $A_{\mu}^{a}$ fields

*The last two terms in (2) are BRST-invariant terms $(K_{\mu}^{a}\delta A_{a}^{\mu} + L_{a}\delta \omega^{a})$ added to the "usual" action for the convenience of work with generating functional $Z[J]$ (see transition from (6.238) to (6.244) through (6.239) and (6.243)); $K_{\mu}^{a}, K_{i}$ are sources for BRST-transformed fields $\delta A_{\mu}^{a}$ and $\delta \omega_{a}$

*$\delta \varphi$ means BRST-transformation of $\varphi$


We have the partition functional
$$\tag{3}
Z[J] = \int D(A, \omega, \dots)e^{i(S + \int \mathrm{d}^{4}x (J_{\mu}^{a}A^{\mu}_{a} + \bar{\eta}^{a}\omega_{a} + \bar{\omega}^{a}\eta_{a}))};
$$
and generating functional for strongly connected diagrams
$$\tag{4}
\Gamma = W - \int \mathrm{d}^{4}x(J_{\mu}^{a}A^{\mu}_{a} + \bar{\eta}^{a}\omega_{a} + \bar{\omega}^{a}\eta_{a}).
$$
The author writes this equation in a chapter about derivation of transverseness of full gauge boson propagator by using Slavnov-Taylor identity.
Question
I don't understand why in $(1)$ cubic functional derivative terms like $[\delta^{3}\Gamma/\delta(\cdots)]$ are zero, and why the terms 
$$\tag{5}\frac{\delta^{2}\Gamma}{\delta A_{\nu}^{c}\delta K_{\mu}^{a}}\frac{\delta^{2}\Gamma}{\delta \omega^{b}\delta A^{\mu , a}}=0.$$ 
Why do they vanish?
 A: One has  $ \dfrac{\delta \Gamma}{\delta K^\mu_a(x)} = sA_\mu^a(x)= \partial_\mu \omega^a(x) - g f^{abc} \omega^b(x)A_{\mu}^ c(x) $
We notice that :  $$sA_\mu^a(x))|_{fields=0}= 0 \tag{1}$$ 
We notice that : 
$$\dfrac{\delta}{\delta \omega^b(y)}sA_\mu^a(x)|_{fields=0} = (\partial_\mu \delta^a_b \delta(x-y) -   g f^{abc} \delta(x-y)A_{\mu}^ c(x)|_{fields=0} = \partial_\mu \delta^a_b \delta(x-y)\tag{2}$$
So this term is not zero.
We notice also that : 
$$\dfrac{\delta}{\delta A_\nu^c(z)}sA_\mu^a(x)|_{fields=0} = - g f^{abc} \omega^b(x) \delta_\mu^\nu\delta (x-y)|_{fields=0} =0  \tag{3}$$
One has : $\dfrac{\delta \Gamma}{\delta A_\mu^a(x)} =  - J^\mu_a(x) $, and, obviously:
$$J^\mu_a(x))|_{sources=0} =0 \tag{4}$$
So, $\dfrac{\delta \Gamma}{\delta K^\mu_a(x)} \dfrac{\delta \Gamma}{\delta A_\mu^a(x)} = - sA_\mu^a(x) J^\mu_a(x) $
Then : 
$\dfrac{\delta}{\delta A_\nu^c(z)}\dfrac{\delta}{\delta \omega^b(y)} (\dfrac{\delta \Gamma}{\delta K^\mu_a(x)} \dfrac{\delta \Gamma}{\delta A_\mu^a(x)})|_{fields=sources=0}\\= \dfrac{\delta}{\delta A_\nu^c(z)}\dfrac{\delta}{\delta \omega^b(y)} (-sA_\mu^a(x) J^\mu_a(x)|_{fields=sources=0} $
The double derivation gives four terms, the first term is :
$$-\dfrac{\delta}{\delta A_\nu^c(z)}\dfrac{\delta}{\delta \omega^b(y)}  (sA_\mu^a(x))|_{fields=sources=0} \quad  J^\mu_a(x)|_{fields=sources=0}\tag{I}$$
and is zero because of $(4)$
The second term is :
$$-\dfrac{\delta}{\delta A_\nu^c(z)}  (sA_\mu^a(x))|_{fields=sources=0} \quad \dfrac{\delta}{\delta \omega^b(y)} J^\mu_a(x)|_{fields=sources=0}\tag{II}$$
and is zero because of $(3)$
The third term is :
$$-\frac{\delta}{\delta \omega^b(y)}   (sA_\mu^a(x))|_{fields=sources=0} \quad \dfrac{\delta}{\delta A_\nu^c(z)}J^\mu_a(x)|_{fields=sources=0}\tag{III}$$
and is different of zero (see $(2)$)
The last term is :
$$- (sA_\mu^a(x))|_{fields=sources=0} \quad \frac{\delta}{\delta \omega^b(y)}  \dfrac{\delta}{\delta A_\nu^c(z)}J^\mu_a(x)|_{fields=sources=0}\tag{IV}$$
and is  of zero because of $(1)$
So, finally, we have only the third term, that you may rewrite : 
$\dfrac{\delta}{\delta \omega^b(y)}\dfrac{\delta}{\delta K^\mu_a(x)}   (\Gamma)|_{fields=sources=0} \quad \dfrac{\delta}{\delta A_\nu^c(z)}\dfrac{\delta}{\delta A_\mu^a(x)}(\Gamma)|_{fields=sources=0}$
[UPDATE]
First, I must correct my previous comment which was wrong. In fact, with the functional $\Gamma(\Phi, J_{BRST})$, where $\Phi$ represent all fields, and $J_{BRST}$ the sources of the BRST variations of fields (the $K, L$), when we calculate vertex functions, we take successive derivatives of $\Gamma(\Phi, J_{BRST})$ relatively to the $\phi$ and/or relatively to the $J_{BRST}$, and after we put $\Phi= J_{BRST}=0$. This is the standard procedure, and it is implicit in most textbooks. Note that you have the same thing for the connected diagram functional $W(J,J_{BRST})$: to obtain correlation functions, you derive  $(J,J_{BRST})$ relatively to the $J$ and/or the $J_{BRST}$ and you make $J= J_{BRST}=0$. All this is standard and "implicit". So, coming back to our vertex expression of derivatives of $\Gamma(\Phi, J_{BRST})$, the fact that we put the fields to zero, is standard, and has nothing to do with the vacuum. The extra-condition is here $J=0$, and it is this condition which is related to the fact that we are considering vaccuum diagrams, and it is why, in your textbook, this condition is explicitely stated. Now, to answer to your first comment, the $J = \frac{\partial \Gamma}{\partial A}$ depends on the $A, \omega$, because the vertex functions like $\frac{\partial \Gamma}{\partial A \partial A \partial \omega}$ are not null, and in fact you could traduce them, with propagators, in correlations functions $\frac{\partial W}{\partial J \partial J \partial \eta}$ which are not zero.
