= Solution
Absorb $k_BT$ into the <free energy>, so the statistical weight is $e^{-F}$. Under the <Gaussian fixed point> rescaling, invariance of the <gradient energy> gives the <engineering dimension>
$$
\boxed{[\phi]=\frac{d-2}{2}=\frac12\quad(d=3)}.
$$
The <engineering dimensions> of $\phi^2,\phi^4,\phi^6$ are respectively $1,2,3$. Their coupling <eigenvalues> are $d-[\phi^{2m}]$, hence $2,1,0$. \b[The quadratic and quartic interactions are <relevant operators>; the sextic interaction is a <marginal operator> at this order.]
The three steps of the <momentum-shell renormalization group> are:
* Split $\phi=\varphi+\eta$ into slow modes $|k|<\Lambda/b$ and fast modes $\Lambda/b<|k|<\Lambda$, and integrate over $\eta$.
* Rescale $x'=x/b$ to restore the <ultraviolet cutoff> to $\Lambda$.
* Rescale the field as $\phi'(x')=b^{(d-2)/2}\varphi(bx')$ to normalize the <gradient energy>.
Here $b=e^s>1$, so increasing $s$ means flowing toward longer distances. At leading order, the fast field has a <Gaussian distribution> with covariance
$$
C_b(x-y)=\int_{\Lambda/b<|k|<\Lambda}\frac{d^3k}{(2\pi)^3}\frac{e^{ik\cdot(x-y)}}{k^2+\mu_0^2},\qquad I_b=C_b(0).
$$
This perturbative integration requires $k^2+\mu_0^2>0$ on the shell. By the <Wick theorem>, odd fast-field moments vanish and $\langle\eta^4\rangle=3I_b^2$, $\langle\eta^6\rangle=15I_b^3$. Therefore the first term of the <cumulant expansion of a coarse-grained free energy> is
$$
\lambda_0\langle(\varphi+\eta)^6\rangle=\lambda_0\left(\varphi^6+15I_b\varphi^4+45I_b^2\varphi^2+15I_b^3\right).
$$
This is the <quartic interaction generated by a sextic interaction>. Including the rescaling, the requested flow is
$$
\boxed{g(b)=15b\lambda_0\!\int_{\Lambda/b<|k|<\Lambda}\frac{d^3k}{(2\pi)^3}\frac{1}{k^2+\mu_0^2}+O(\lambda_0^2)}.
$$
In spherical coordinates, $I_b=(2\pi^2)^{-1}\int_{\Lambda/b}^{\Lambda}k^2/(k^2+\mu_0^2)\,dk$. In particular, for $\mu_0^2=0$,
$$
g(b)=\frac{15\lambda_0\Lambda}{2\pi^2}(b-1)+O(\lambda_0^2),\qquad \frac{dg}{ds}=g+\frac{15\Lambda}{2\pi^2}\lambda+\cdots.
$$
The same integration shifts the quadratic coefficient by $90\lambda_0 I_b^2$ before rescaling and adds an irrelevant constant to the <free energy>. Thus setting the bare quartic coupling to zero does not place the theory on the tricritical <critical surface>.
To calculate the sextic <renormalization-group beta function> to second order, let $V=\lambda_0\int d^3x\,(\varphi+\eta)^6$. The required <cumulant expansion> is
$$
F_{\mathrm{eff}}=F_0+\langle V\rangle-\frac12\left(\langle V^2\rangle-\langle V\rangle^2\right)+O(\lambda_0^3).
$$
The subtraction removes disconnected contributions. The term $20\lambda_0\varphi^3\eta^3$ at each vertex supplies the primitive sextic correction. The <Wick contractions> give
$$
\langle\eta(x)^3\eta(y)^3\rangle=6C_b(x-y)^3+9I_b^2C_b(x-y).
$$
The first term contains three propagators connecting the two vertices; its factor $6$ is the number of bijections between the three fields at each vertex. The second contains a local <tadpole diagram> at each vertex and one connecting propagator. The primitive contribution to the <free energy> is therefore
$$
\Delta F_6=-1200\lambda_0^2\int d^3x\,d^3y\,\varphi(x)^3\varphi(y)^3C_b(x-y)^3.
$$
Expanding the slow fields about a common point gives a negative local sextic correction proportional to
$$
J_b=\int d^3z\,C_b(z)^3=\int\frac{d^3p\,d^3q}{(2\pi)^6}\frac{\chi_b(p)\chi_b(q)\chi_b(p+q)}{(p^2+\mu_0^2)(q^2+\mu_0^2)((p+q)^2+\mu_0^2)}\geq0,
$$
where $\chi_b$ restricts a momentum to the shell. This <connected Feynman diagram> has two independent loop momenta. Other <Wick contractions> also generate lower interactions and <tadpole diagrams>; they must be included or subtracted consistently when fixing the tricritical <critical surface>.
To extract the logarithmic <tricritical sextic beta function>, use <normal ordering> to remove local tadpoles and tune the quadratic and quartic <relevant operators>. The massless three-dimensional propagator is $C(z)=1/(4\pi|z|)$. The same three-line contraction over short separations $a<|z|<ba$ gives
$$
\int_{a<|z|<ba}d^3z\,C(z)^3=\frac{1}{16\pi^2}\log b.
$$
Its positive coefficient and the negative sign of the second <cumulant expansion> term establish the sign of the flow. Matching this logarithm to a renormalized coupling, or performing consistent iterated shell integration, gives
$$
\boxed{\frac{d\lambda}{ds}=-C\lambda^2+O(\lambda^3),\qquad C>0},\qquad \lambda(s)=\frac{\lambda(0)}{1+C\lambda(0)s}+\cdots.
$$
The precise normalization of $C$ is not needed. In a sharp momentum cutoff scheme, simply forcing all three internal momenta into one infinitesimal shell does not extract this two-loop logarithm: the shell restrictions remove its linear-in-$ds$ phase space. The finite-shell cumulant and the logarithmic renormalized calculation above are distinct stages of the computation. Field normalization corrections at order $\lambda^2$ enter this <renormalization-group beta function> only at order $\lambda^3$.
\b[A positive, stabilizing sextic coupling is <marginally irrelevant> in three dimensions on the tricritical <critical surface>.] Reversing the convention for $s$ reverses the sign of the <renormalization-group beta function>. A negative sextic coupling without higher stabilizing powers makes the potential unbounded; moreover, the untuned bare choice $g=0$ generally flows away through its generated <relevant operator>.
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