= Solution
Set the renormalized $g_3$ to zero. For a six-scalar vertex, <3/b> gives $\omega=-n_2$, so a new overall <logarithmic divergence> can only occur in graphs with $n_2=0$. With $b=6,f=0$, the line counts become $I_f=n_1$ and $I_b=n_1-3$, hence $L=n_1-2$.
At one loop this forces $n_1=3$ and $I_b=0$: the only candidate is a closed <triangle Feynman diagram> with two external scalar legs at each vertex. Its numerator at large <loop momentum> is proportional to
$$
\operatorname{tr}[(\not k)^3]=k^2\operatorname{tr}(\not k)=0.
$$
This vanishes for the specified two-component <gamma matrices>, each of which is traceless. Replacing one leading $\not k$ by the fermion mass or an external <momentum> lowers the <superficial degree of divergence> from zero to at most $-1$. Thus the apparent logarithm cancels. Graphs with a $g_2$ vertex already have negative overall degree; the pure-scalar one-loop triangle, for instance, has $\omega=-3$. There is no one-loop <ultraviolet subdivergence> inside these proper one-loop vertices. Therefore
$$
\boxed{\delta g_3^{\text{one loop}}=0.}
$$
The six-point <correlation function> need not vanish: finite terms are allowed. Nor is it correct to say that all odd <gamma-matrix traces> vanish in $2+1$ dimensions. In this representation,
$$
\gamma^0\gamma^1\gamma^2=-iI_2,\qquad\operatorname{tr}(\gamma^0\gamma^1\gamma^2)=-2i\ne0.
$$
The cancellation above follows from repeating the same leading <loop momentum>, not from a false four-dimensional trace rule.
At two loops, $n_1=4$, $I_f=4$ and $I_b=1$ are possible. A closed fermion box with a scalar line joining two of its vertices leaves six external scalar legs. It has $L=2$, $\omega=0$, and four leading <gamma matrices>; the one-loop odd-numerator argument no longer applies.
A background-field check demonstrates that its logarithm is actually nonzero. Put $\phi=v+\eta$ with constant $v$. Define the background fermion mass $\mu=m+g_1v^2/2$, the background scalar squared mass $\mathcal B^2=M^2+g_2v^2/2$, and the induced three-leg <Yukawa coupling> $y=g_1v$. The two-loop <quantum effective potential> contains the <sunset diagram> with two fermion lines and one scalar line. After a <Wick rotation>, its relevant integral, apart from a nonzero overall combinatorial factor, is
$$
y^2\int\frac{d^3k\,d^3l}{(2\pi)^6}\frac{2(-k\cdot l+\mu^2)}{(k^2+\mu^2)(l^2+\mu^2)((k-l)^2+\mathcal B^2)}.
$$
Here the dots and <gamma-matrix traces> are Euclidean. With $A=k^2+\mu^2$, $B=l^2+\mu^2$, $C=(k-l)^2+\mathcal B^2$, its numerator is exactly
$$
2(-k\cdot l+\mu^2)=-A-B+C+4\mu^2-\mathcal B^2.
$$
The first three terms reduce to products of one-loop <tadpole integrals>, which have no logarithmic pole near three dimensions in <dimensional regularization>. The remaining term is
$$
y^2(4\mu^2-\mathcal B^2)H,\qquad H=\int\frac{d^3k\,d^3l}{(2\pi)^6}\frac1{ABC}.
$$
For positive regulator masses, the large-momentum region of $H$ has a positive, nonzero logarithmic divergence. One can see this directly by restricting the relative angles and the ratio $|k|/|l|$ to compact intervals away from zero: the remaining common radial integral is proportional to $\int^\Lambda dr/r$. Its proper one-loop subgraphs are ultraviolet convergent.
Finally, the coefficient multiplying $H$ contains
$$
y^2(4\mu^2-\mathcal B^2)=g_1^2v^2\left[4m^2- M^2+\left(4mg_1-\frac{g_2}{2}\right)v^2+g_1^2v^4\right].
$$
The <two-loop sextic counterterm in quartic Yukawa theory> follows because its $v^6$ coefficient is $g_1^4\ne0$ for a nonzero coupling. Other two-loop vacuum topologies reduce to products of finite-in-dimensional-regularization tadpoles and do not cancel this logarithmic pole. Thus
$$
\boxed{\delta g_3^{\text{two loops}}\ \text{is generically nonzero and proportional to }g_1^4.}
$$
A sextic <counterterm> is required even if the renormalized sextic coupling is set to zero at a chosen <renormalization scale>. A theory constrained to $g_3=0$ at every scale lacks <counterterm closure> when $g_1\ne0$. The decoupled exception $g_1=0$ is the three-dimensional scalar <quartic interaction>, which needs no sextic <counterterm> by <power counting>.
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