Consider the typical WZ model with global symmetry $U(1)\times U(1)_R$. Usually we write the superpotential $W$ as

$$W = \frac{M}{2} \Phi^2 + \frac{\lambda}{3} \Phi^3$$

but you have simply done some rescaling, which is ok. I am writting it like this to agree with the usual convention. Then, we require that the, so to be, renormalized superpotential is still a holomorphic function and transforms correctly under the global transformations I quote above. For these requirements to hold the superpotential takes the form

$$ W = \frac{M}{2} \Phi^2 \,\, f \left( \frac{\lambda \Phi^3}{3} \right) $$

Now what happens in the weak coupling limit of the theory? The quantum superpotential should be approaching the classical one, so Taylor expand the function $f$ as

$$ W = \frac{M}{2} \Phi^2 \,\, \left( 1 + \frac{2}{3}\frac{\lambda \Phi}{m} + \text{higher corrections} \right) = \frac{M}{2} \Phi^2 \,\, f \left( \frac{\lambda \Phi^3}{3} \right)+ \text{higher corrections} $$ Finally, the superpotential should be well behaved as $M \to 0$ constraining W to not having negative powers of $M$. It is obvious then that no higher terms are generated at quantum level and thus no counterterms to absorb any divergences of $M$ and $\lambda$. This is also the reason for the phrase "holomorphicity puts constraints in renormalization". Most of these stuff can be found in Terning's book "Modern Supersymmetry: Dynamics and Duality"