= Solution
Let $C(u,v)=2D_F(u-v)$ be the action-consistent free <propagator> established in <4/a>. Write $S=S_0+S_I$ and put
$$
V(z)=\phi(z)^2\phi^*(z)^2,\qquad
S_I=-\frac\lambda4\int d^Dz\,V(z).
$$
The perturbative expansion is
$$
e^{iS}=e^{iS_0}\sum_{n=0}^\infty\frac{(-i\lambda/4)^n}{n!}
\int\prod_{r=1}^n d^Dz_r\,\prod_{r=1}^nV(z_r).
$$
Divide numerator and denominator by the same free <vacuum functional> $Z_0$. For $\mathcal O=\Phi(y_1)\cdots\Phi(y_N)$, they become
$$
\begin{aligned}
\mathcal N[\mathcal O]&=\sum_{n\geq0}\frac{(-i\lambda/4)^n}{n!}
\int\prod_r d^Dz_r\,\left\langle T\mathcal O\prod_r V(z_r)\right\rangle_0,\\
\mathcal D&=\sum_{n\geq0}\frac{(-i\lambda/4)^n}{n!}
\int\prod_r d^Dz_r\,\left\langle T\prod_r V(z_r)\right\rangle_0.
\end{aligned}
$$
For $\mathcal O=\phi(y_1)\phi^*(y_2)$, abbreviate $C_{12}=C(y_1,y_2)$ and $C_{zz}=C(z,z)$. Since only opposite-charge fields contract, <Wick's theorem> gives
$$
\langle V(z)\rangle_0=2C_{zz}^2,
$$
and
$$
\langle T\phi(y_1)\phi^*(y_2)V(z)\rangle_0
=2C_{12}C_{zz}^2+4C(y_1,z)C_{zz}C(z,y_2).
$$
The first multiplicity is $2!$ internal pairings. In the connected term, there are two choices for the vertex's $\phi^*$ contracted to the external $\phi$ and two for its $\phi$ contracted to the external $\phi^*$; the remaining pair forms a <tadpole diagram>. Therefore
$$
\begin{aligned}
\mathcal N&=C_{12}
-\frac{i\lambda}{2}C_{12}\int d^Dz\,C_{zz}^2
-i\lambda\int d^Dz\,C(y_1,z)C_{zz}C(z,y_2)+O(\lambda^2),\\
\mathcal D&=1-\frac{i\lambda}{2}\int d^Dz\,C_{zz}^2+O(\lambda^2).
\end{aligned}
$$
Expanding the inverse denominator cancels the free-line times <vacuum bubble> term, leaving
$$
\boxed{\langle T\phi(y_1)\phi^*(y_2)\rangle
=C_{12}-i\lambda\int d^Dz\,C(y_1,z)C(z,z)C(z,y_2)+O(\lambda^2).}
$$
The local quartic <interaction vertex> has factor $-i\lambda$, since $2!2!$ cancels its denominator 4. The oriented <tadpole diagram> has <Feynman-diagram symmetry factor> 1; the isolated two-loop vacuum vertex has factor $1/2$ in its coefficient. Coincident kernels and the integrals require a common <regularization>; none is being assigned a finite numerical value here.
\Image[/past-exam-of-the-quantum-field-theory-course-of-the-university-of-oxford/2018-complex-scalar-two-point-diagrams.png]
{title=Free propagation, its disconnected vacuum factor, and the connected oriented tadpole correction}
{description=An arrow from u to v represents C(u,v). The disconnected vacuum factor cancels against the denominator of the normalized correlation function.}
More generally, <vacuum bubbles> exponentiate and factor from every numerator. Here <global U(1) symmetry> also makes all one-point functions zero. Thus a surviving graph with the two external insertions in separate connected components would have a component with one unbalanced charge and vanish. This proves that only connected graphs contribute to this particular two-point function to all perturbative orders. Vacuum normalization alone would not remove disconnected components containing external legs in a general four-point function. If one formally chooses a different free kernel, the displayed Wick multiplicities are unchanged; replacing $C$ by $C_{\mathrm{lit}}$ gives a formal oscillator series, not the Gaussian integral of the action as printed.
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