Deriving a scattering amplitude of a loop diagram using path integral formulation I am following Zee's QFT in a nutshell and his excerise I.7.2 states to derive
$$\frac{1}{2}(-i\lambda)^2 \int \frac{d^4k}{(2 \pi)^4}\frac{i}{k^2 -m^2 +i\epsilon}\frac{i}{(k_1+k_2-k)^2 -m^2 +i\epsilon}\tag{I.7.23}$$
from the first principles, that is from
$$Z(J) = \int D\phi e^{i \int d^4x \frac{1}{2}[(\partial \phi)^2 - m^2\phi^2] -(\lambda/4!)\phi^4 + J\phi}.\tag{I.7.11}$$
The amplitude written is just a basic loop diagram of order $\lambda^2$.
My attempt is to follow the Zee's approach in which i expand $Z(J)$ in series of $\lambda$ and $J$. So right before the expansion in the series I have
$$Z(J) = Z(0,0) e^{-(i/4!)\lambda\int d^4w \left (\frac{\delta}{i\delta J(w)}\right)^4}e^{-(i/2)\iint d^4x d^4y J(x) D(x-y)J(y)}$$
As i want 2 vertices and 6 lines, in the series expansion I choose $\lambda^2$ and $J^6$ and I have
$$Z(J) = Z(0,0) \frac{1}{2!}\left[\frac{-i}{3!}\lambda \int d^4w \left( \frac{\delta}{i\delta J(w)}\right)\right]^2\frac{1}{6!}\left[\frac{-i}{2}\iint d^4xd^4y J(x)D(x-y)J(y)\right]^6.$$
After some algebra i get
$$Z(0,0)\frac{i^8\lambda^2}{2^73!3!6!}\iint d^4w d^4z D(w-z)D(w-z)D(x_1-w)D(x_2-w)D(x_3-z)D(x_4-z)$$
And this, to me at least, does look like a loop diagram. We have 2 inner lines $D(w-z)$ and 4 external $D(x_i - w(z))$. But here is where my understanding stops. I dont know how to get from this to the amplitude itself. This part is not really covered in the Zee's book that well. I would appreciate if a bit more detailed explanation is given.
 A: I was writing this as a comment, but it got too long so I'm including it as an answer instead.
You might want to go through the calculation again, the term of interest in Z[J] before taking any derivatives to extract the 4-pt function should have a factor of:
$$
\frac{1}{16}\quad \textrm{instead of what you quote},\quad \frac{1}{2^73!3!6!}
$$
See in particular the $\lambda^2$ coefficient in $W[J]$ in eqn 2.16 in arxiv.org/abs/1512.02604. The quantity $W[J]$ is as usual the exponentiated version of $Z[J]$, so that it generates all connected and renormalised Green functions. Once you have the correct combinatorial factor you need the 4th $J$ derivative and then you need to go to momentum space by Fourier transforming all position space Green functions.
Slightly more precisely, the 4-pt 1-loop position-space Green function is (schematically):
$$
G\sim \delta_J^4 Z[J]\,\big|_{J=0},
$$
and the relevant term in $Z[J]$ is:
$$
Z[J] = \dots +\frac{1}{2}\Big(\frac{\lambda}{4!} \int \delta_J^4\Big)^2 Z_0[J]+\dots,
$$
where $Z_0[J]$ is the $\lambda=0$ version of $Z[J]$, so that $Z_0[J]$ is Gaussian in $J$. All in all, the derivative you need to calculate the 4-pt one-loop amplitude is:
$$
G\sim \delta_J^4\frac{1}{2}\Big(\frac{\lambda}{4!} \int \delta_J^4\Big)^2 Z_0[J]\,\big|_{J=0}.
$$
In fact, there's also a tadpole (or 'cephalopod') diagram at the same order (see the diagram immediately after the one of interest in eqn (2.16) in  arxiv.org/abs/1512.02604) so in your derivation you will find the associated extra terms in addition to the diagram of interest.
