Introductory overview
I have that
$$iW_0[J] := -\frac{1}{2}\int d^4x d^4 y J(x)D_F(x-y)J(y)$$
and I'm trying to perform the calculation of a two-point function $G^{(2)}(x,y)$ from the fully connected generating functional $W[J]$ defined by $Z[J] = e^{iW[J]}$. The interaction is the phi-four theory. The $iW[J]$ is given by the formula
$$iW[J] = iW_0[J] + \ln\left(1 + \underbrace{e^{-iW_0[J]}\left(e^{-i\frac{\lambda}{4!}\int d^4 x \frac{\delta^4}{\delta J(x)^4}}-1\right)e^{+iW_0[J]}}_{A}\right)$$
I need to calculate the two-point function up to order $\lambda$ only, so in the path integral formalism we get
$$G^{(2)}(x,y) = \frac{(-i)^2}{iW[0]}\frac{\delta^2}{\delta J(x)\delta J(y)}\left(1 - \frac{i\lambda}{4!}\int d^4z \frac{\delta ^4}{\delta J(z)^4}\right)iW[J]$$
Were we made the change $Z[J] \to iW[J]$ in the formulas to get only connected terms. In order to calculate this up to order $\lambda$ I think that it is needed $A$ up to order $\lambda$...
Question
My question is how can I calculate this term up to order $\lambda$: The main problem is in calculating the term
$$\frac{\delta ^4}{\delta J(z)^4}e^{iW_0[J]} = \dots$$
The first term would be like
$$\frac{\delta}{\delta J(z)} e^{iW_0[J]} = -\frac{1}{2}\int d^4x J(x)D_F(x-z)e^{iW_0[J]}$$
but this is only a choice of were I performed the derivative. I'm using the rule
$$\frac{\delta}{\delta J(z)} e^{i \int d^4 y J(y)\phi(y)} = i\phi(z)e^{i \int d^4 y J(y)\phi(y)}$$
And worse, after some of the calculations we get terms like $D_F(0)$, and what should I do with them?
I'm not sure that I get your question correctly, but when you take the functional derivative of $$iW_0[J] := -\frac{1}{2}\int d^4x \, d^4y \, J(x) \, D_F(x-y) \,J(y)$$ you should take the derivative of each occurrence of $J$, like in the product rule $(uv)' = u'v + uv'$: $$ \frac{\delta(iW_0[J])}{\delta J(z)} = - \frac{1}{2} \frac{\delta}{\delta J(z)} \int d^4x \, d^4y \, J(x) \, D_F(x-y) \,J(y) \\ = - \frac{1}{2} \left( \int d^4x \, d^4y \, \delta(x-z) \, D_F(x-y) \, J(y) + \int d^4x \, d^4y \, J(x) \, D_F(x-y) \, \delta(y-z) \right) \\ = - \frac{1}{2} \left( \int \, d^4y \, D_F(z-y) \, J(y) + \int d^4x \, J(x) \, D_F(x-z) \right) \\ = - \int \, d^4y \, \frac{D_F(z-y)+D_F(y-z)}{2} \, J(y) $$