Solution
= Solution
The scalar has canonical dimension $[\phi]=(d-2)/2$, so
$$
[\lambda]=d-6\frac{d-2}{2}=6-2d.
$$
The $\phi^6$ coupling is therefore marginal in $d=3$. Its leading two-point correction is the order-$\lambda$ diagram with one six-leg vertex, two external legs, and the remaining four legs paired into two tadpole loops. This diagram is independent of external momentum, so it renormalizes the mass but has no $p^2$ pole and does not contribute to $\delta_Z$.