= Solution
For a <chiral superfield> of <mass dimension> one, a <renormalizable quantum field theory> has a quadratic <Kähler potential> and a <superpotential> of degree at most three. Before canonical normalization one may write
$$
K=Z\Phi^\dagger\Phi+[k_1\Phi+k_2\Phi^2+\mathrm{h.c.}]+k_0,\qquad Z>0,
$$
with real $k_0$, and
$$
W=w_0+\ell\Phi+\frac m2\Phi^2+\frac g3\Phi^3.
$$
The purely holomorphic terms in $K$, their conjugates and its constant have vanishing full <superspace integration>; they are <Kähler transformations> of the global theory. Rescaling the <chiral superfield> sets $Z=1$, so the physically general kinetic choice is $K=\Phi^\dagger\Phi$ up to these terms. The <constant superpotential in global supersymmetry> also has no component effect. The dimensions of $\ell,m,g$ are respectively two, one and zero.
To eliminate the linear term, translate the complex <chiral superfield> as $\Phi=\Phi'+s$. The shifted linear coefficient is
$$
W'(s)=\ell+ms+gs^2.
$$
If $g\ne0$, this quadratic has a root over the complex numbers; if $g=0$ but $m\ne0$, take $s=-\ell/m$. The resulting linear term vanishes and the new quadratic coefficient is $m+2gs$. The canonical <Kähler potential> changes only by holomorphic, antiholomorphic and constant terms, which do not change the global kinetic action. This is the <superpotential critical-point shift>.
\b[The claim needs an exception:] if $m=g=0$ but $\ell\ne0$, the <superpotential> is purely linear and no invertible translation can remove it. More generally an invertible regular field redefinition cannot turn a nowhere-zero derivative into a critical point. The theory with $W=\ell\Phi$ has $F=-\bar\ell$, a constant positive <scalar potential> $|\ell|^2$, and <flat F-term breaking with a linear superpotential>. Below, $m$ and $g$ denote the quadratic/cubic coefficients after the valid shift, as intended by the question.
Expand $\Phi=\phi+\sqrt2\theta\psi+\theta^2F$ in chiral coordinates. The highest component of $W(\Phi)$ is $W'(\phi)F-\tfrac12W''(\phi)\psi\psi$, while the full-superspace kinetic term contains $F^*F$. Thus the <auxiliary field> part of the component <Lagrangian> is
$$
\mathcal L_F=F^*F+(m\phi+g\phi^2)F
+(\bar m\bar\phi+\bar g\bar\phi^2)F^*.
$$
Varying $F$ and $F^*$ independently gives
$$
F^*=-(m\phi+g\phi^2),\qquad F=-(\bar m\bar\phi+\bar g\bar\phi^2).
$$
Substitution gives $\mathcal L_F=-|m\phi+g\phi^2|^2$, and hence the <F-term scalar potential>
$$
\boxed{V(\phi)=|m\phi+g\phi^2|^2.}
$$
It is nonnegative and vanishes at $\phi=0$, and also at $\phi=-m/g$ when $g\ne0$. These are <supersymmetric vacua>, since the <auxiliary field> vanishes. Thus \b[the quadratic/cubic model does not spontaneously break supersymmetry]. If $m=0$ but $g\ne0$, zero is the sole critical point; if both vanish, every $\phi$ is a zero-energy vacuum. Nonzero $m,g$ give the usual two vacua, coinciding when $m=0$.
For the loop argument, promote the coefficients to background chiral <spurions>. One convenient pair of independent formal $U(1)$ charge assignments is
$$
\begin{array}{c|rrrr|r}
&\Phi&m&g&\theta&W\\\hline
U(1)&1&-2&-3&0&0\\
U(1)_R&2&-2&-4&1&2
\end{array}.
$$
Both symmetries act nontrivially on all three of $\Phi,m,g$; the second is an <R-symmetry>. The <superspace integration> measure $d^2\theta$ has <R-charge> $-2$, so both tree monomials give invariant actions. These are spurionic selection rules, not necessarily actual symmetries after fixing numerical couplings. The R-row is an ordinary-charge redefinition of the usual <Wess–Zumino spurion charge assignment>.
A local Wilsonian <superpotential> is a <holomorphic function> of the chiral fields and chiral <spurions>. The ratio $z=g\Phi/m$ is neutral under both charge assignments and has dimension zero, whereas $m\Phi^2$ has exactly the charges and dimension of $W$. The general symmetry-allowed form, on a patch with $m\ne0$ and up to a physically irrelevant global constant, is therefore
$$
\boxed{W_{\mathrm{eff}}=m\Phi^2 f\left(\frac{g\Phi}{m}\right),}
$$
with $f$ holomorphic where that description is regular. Symmetry by itself does not determine $f$.
For a regular perturbative local <Wilsonian effective action> retaining the same elementary field and an infrared cutoff, expand $f(z)=\sum_n c_nz^n$. Its term of order $n$ is $c_n g^n m^{1-n}\Phi^{n+2}$. Negative $n$ are singular in $g$, while $n\geq2$ are singular in $m$; these are excluded by perturbative holomorphic regularity at the free-coupling limits. Thus only the quadratic and cubic terms survive. Their regular holomorphic coefficients cannot acquire coupling-dependent invariant corrections. At $g=0$ the quadratic theory is Gaussian, fixing $c_0=1/2$ exactly. A contribution linear in $g$ to the cubic vertex has only one cubic interaction: three external lines exhaust that vertex, leaving no contraction to form a loop. Thus $c_1=1/3$ as well in the present normalization. This supplies the <holomorphy argument for superpotential non-renormalization> rather than concluding non-renormalization from symmetry alone.
The <Kähler potential> can still undergo <wave-function renormalization>. If its quadratic coefficient becomes $Z$, canonical normalization gives $m_c=m/Z$ and $g_c=g/Z^{3/2}$, so physical masses and interactions can run despite <superpotential non-renormalization>. More generally a positive nonsingular effective <Kähler metric> replaces $V$ by $K^{\phi\bar\phi}|W'(\phi)|^2$ and does not remove its zero-energy critical points. Infrared-singular nonlocal one-particle-irreducible terms and eliminating the entire massive field are outside this Wilsonian statement.
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