= Solution
Define the free <Feynman propagators> by
$$
D_0(x)=\int\frac{d^4k}{(2\pi)^4}\frac{i\,e^{-ik\cdot x}}{k^2+i0},\qquad
D_m(x)=\int\frac{d^4k}{(2\pi)^4}\frac{i\,e^{-ik\cdot x}}{k^2-m^2+i0}.
$$
The interaction preserves the global phase symmetry $\varphi\mapsto e^{i\alpha}\varphi$, with the real $\phi$ neutral. Therefore the charge-two correlator vanishes in a symmetry-invariant perturbative vacuum:
$$
\boxed{\langle T\varphi(x)\varphi(y)\phi(z)\rangle=0.}
$$
In <Wick contractions>, a complex field contracts only with its conjugate; one vertex has one of each and cannot absorb both external $\varphi$ fields. The symmetry shows this vanishing holds to all perturbative orders, not only first order.
The neutral connected correlator has one allowed contraction to each external point at order $g$:
$$
\boxed{G_{\rm conn}(x,y,z)=-ig\int d^4w\,
D_0(x-w)D_0(y-w)D_m(z-w)+O(g^3).}
$$
The propagators are even functions of their separation; their orientations encode the conjugate-field contractions. Equivalently, with every external momentum Fourier transformed as incoming,
$$
G_{\rm conn}(x,y,z)=(-ig)\int\frac{d^4k\,d^4\ell\,d^4r}{(2\pi)^{12}}
(2\pi)^4\delta^{(4)}(k+\ell+r)e^{-i(kx+\ell y+rz)}
\frac{i}{k^2+i0}\frac{i}{\ell^2+i0}\frac{i}{r^2-m^2+i0}+O(g^3).
$$
No factor of two accompanies the vertex: $\varphi$ and $\varphi^\dagger$ are distinct fields.
\b[For the full correlator as literally written, a <disconnected scalar three-point tadpole> is also possible.] Contract the two external complex fields with each other and contract the complex pair at the vertex into a loop. Unless <normal ordering> is imposed on the interaction or a <tadpole counterterm> fixes $\langle\phi\rangle=0$, add
$$
\boxed{G_{\rm disc}(x,y,z)=D_0(x-y)\left[-igD_0(0)\int d^4w\,D_m(z-w)\right]+O(g^3).}
$$
The coincident propagator is regulator dependent; in <dimensional regularization> the massless scaleless <tadpole diagram> vanishes. More generally this is $\langle T\varphi^\dagger(x)\varphi(y)\rangle_0\langle\phi(z)\rangle$ to this order and is removed by taking the connected correlator or imposing the stated vacuum-renormalization condition. Normalizing by vacuum bubbles alone does not remove an external-leg disconnected <tadpole diagram>. The charged correlator remains zero in every one of these conventions.<Quantum Field Theory by David Tong>[§§3.3.2 and 3.7.1, Wick expansion, connected Green functions and vacuum normalization]{ref}
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