= Solution
Keep the same measure normalization as in the free <generating functional> and define
$$
W[J]=\mathcal N\int_{\rm periodic}\mathcal Dx\,
\exp\left[-S_0[x]-\int_0^\beta d\tau\left(\frac\lambda{4!}x^4+\frac\mu{6!}x^6\right)+\int_0^\beta d\tau\,Jx\right].
$$
Here $W$ is the source-dependent <partition function>, despite the common alternative convention in which that letter denotes its logarithm. A <functional derivative> of $e^{\int Jx}$ brings down one factor $x(\tau)$ at the differentiated point. Expanding the interaction exponential and replacing each such factor by $\delta/\delta J(\tau)$ therefore gives
$$
\boxed{W[J]=\exp\left[-\int_0^\beta d\tau\left\{\frac\lambda{4!}\left(\frac\delta{\delta J(\tau)}\right)^4+
\frac\mu{6!}\left(\frac\delta{\delta J(\tau)}\right)^6\right\}\right]W_0[J].}
$$
Completing the square in the free <Gaussian integral> gives $W_0[J]=W_0[0]\exp(\tfrac12\int JG^{(0)}J)$ because $G^{(0)}=D^{-1}$ with periodic <boundary conditions>. Restoring $\hbar$ replaces the upper limit by $\hbar\beta$ and the interaction coefficients by $\lambda/(4!\hbar)$ and $\mu/(6!\hbar)$, exactly as in the source.
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