= Solution
Keep $Z=e^{iW}$, define the mean <scalar field> $\varphi(x)=\delta W/\delta J(x)$, and locally invert the source-field relation. The <quantum effective action> is the <Legendre transform>
$$
\boxed{\Gamma[\varphi]=W[J]-\int d^dx\,J(x)\varphi(x),\qquad\frac{\delta\Gamma}{\delta\varphi(x)}=-J(x).}
$$
Its value at zero source is expanded around $\varphi_0=\langle\phi\rangle$, which need not vanish. The kernels $\Gamma^{(n)}=\delta^n\Gamma/\delta\varphi^n|_{\varphi_0}$ are the <one-particle-irreducible vertices>, with the Minkowski diagram vertex factor $i\Gamma^{(n)}$. A <one-particle-irreducible Feynman diagram> remains connected after any one internal line is cut. <Amputation of external propagators> removes the full two-point functions on its external legs. For higher-point connected graphs amputation alone need not remove internal <one-particle-reducible Feynman diagrams>; the derivatives of $\Gamma$ select precisely the irreducible kernels.
To derive the three-point relation, use condensed indices that include spacetime integration. Put $D_{ij}=W_{,ij}=iG_{2,ij}^c$ and $K=\Gamma^{(2)}$. Differentiating $\Gamma_{,i}=-J_i$ gives $K_{ia}D_{aj}=-\delta_{ij}$, so $K=-D^{-1}$. A further derivative, using $\delta D_{ab}/\delta\varphi_k=W_{,ab\ell}(D^{-1})_{\ell k}$ and the derivative of an inverse, gives
$$
\Gamma^{(3)}_{ijk}=(D^{-1})_{ia}(D^{-1})_{jb}(D^{-1})_{kc}W_{,abc}.
$$
Since $W^{(3)}=-G_3^c$ and $D^{-1}=-i(G_2^c)^{-1}$, the <three-point amputation in the connected-action convention> is
$$
\boxed{i\Gamma^{(3)}_{ijk}=\int da\,db\,dc\,(G_2^c)^{-1}_{ia}(G_2^c)^{-1}_{jb}(G_2^c)^{-1}_{kc}G_{3,abc}^c.}
$$
Equivalently, attach a full <two-point correlation function> to each leg of $i\Gamma^{(3)}$ to reconstruct $G_3^c$. The connected functions are evaluated at the same zero-source background. The displayed factors of $i$ follow from the declared Minkowski convention; Euclidean generating functionals redistribute these factors.
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