What Lagrangian counterterms might be needed in a $\phi^ 6$ theory in 3D? Assuming a massive $\phi^6$ theory in $d=3$ given by the Lagrangian
$$\mathcal L=\frac{1}{2}(\partial\phi)^2-\frac{1}{2}m^2\phi^2-\frac{\lambda}{6!}\phi^6 +\mathcal L_\text{ct},$$
what are the counter terms $\mathcal L_\text{ct}$ we need to renormalise it? I think there ought to be four: a tadpole, a mass, a partial derivative one and an interaction one. But I am not sure how you would write these.
 A: Let's work with $\hbar = c = 1$, such that the units of relevant physics quantities can be expressed in terms of mass. I denote the dimension of a quantity by the number of powers of mass. E.g., if $m$ has dimension mass, $[m] = 1$, $[m^2]=2$ etc.
First, let's check your theory is renormalizable. We'll do it slightly naively by checking the the couplings in front of the operators don't have a negative mass dimension. We can find the dimension of the field $\phi$ by noting that the action 
$$
S = \int d^dx \mathcal{L}
$$
must be dimensionless as it appears as $e^{iS}$, $[S]=0$. This allows us to infer that $[\mathcal{L}] = d$. Since $[\partial]= 1$, from the kinetic term we deduce that $[\phi] = (d - 2) / 2$.
Now, if we write a general operator as $\mathcal{L} \supset c \phi^n$, from the requirement that that the operator is superficially renormalizable, $[c] \ge 0$, you can readily show that 
$$
n \le 2 d / (d - 2).
$$
You have $d=3$, so must have $n \le 6$, which you do. 
Let's check your Lagrangian includes all the renormalizable operators. First, note that you are missing a $\phi^4$ operator in your Lagrangian. This one is needed as it will anyway be generated from the $\phi^6$ one. You don't need to add the odd-power operators, e.g. $\phi$ or $\phi^3$, as without them you have a $\mathbb{Z}_2$ discrete symmetry, i.e., your Lagrangian is invariant under $\phi\to-\phi$, so they can never be generated in perturbation theory.
Finally, let's consider the counterterm Lagrangian. Well, by definition of a renormalizbale theory, the counterterm Lagrangian has the same form as the original Lagrangian! but the operators come with different coefficients. The finite parts of the coefficients depend on the choice of renormalization scheme.  
