= Solution
For a single <chiral superfield>, the <supergravity F-term potential>, with $\kappa=m_p^{-1}$ restored, is
$$
V=e^{\kappa^2K}\left(K^{z\bar z}|D_zW|^2-3\kappa^2|W|^2\right),
\qquad D_zW=W_z+\kappa^2K_zW.
$$
The canonical <Kähler potential> has $K_{z\bar z}=K^{z\bar z}=1$, so the <Kähler covariant derivative of a superpotential> is
$$
D_zW=\mu m_p\left[1+\kappa^2\bar z(z+\beta)\right].
$$
Consequently the <Polonyi model> has the <scalar potential>
$$
\boxed{V=|\mu|^2m_p^2e^{\kappa^2|z|^2}
\left(\left|1+\kappa^2\bar z(z+\beta)\right|^2
-3\kappa^2|z+\beta|^2\right).}
$$
The negative term is essential; retaining only the square of the <supergravity auxiliary field> would give the global <F-term scalar potential> rather than the <supergravity F-term potential>.
Set $x=\kappa z=a+iy$ and $b=\kappa\beta=2-\sqrt3$. In this notation $V=|\mu|^2m_p^2e^{|x|^2}U$, where
$$
U=\left(1+a^2+y^2+ba\right)^2+b^2y^2
-3\left[(a+b)^2+y^2\right].
$$
At $a_0=\sqrt3-1$, $y=0$, one has $a_0+b=1$ and $1+a_0(a_0+b)=\sqrt3$, so $U=0$. To prove that this is a <local minimum> in both real directions, rather than checking only the real axis, put $u=a-a_0$. Direct expansion gives the <stable zero-energy Polonyi vacuum> identity
$$
U=\left(u^2+y^2+\sqrt3u\right)^2
+(2\sqrt3-3)u^2+(4-2\sqrt3)y^2.
$$
Every coefficient outside the square is strictly positive, and the exponential factor is positive. Thus $V\ge0$ everywhere and equality requires $u=y=0$, provided $\mu\ne0$. This proves the stronger conclusion
$$
\boxed{\langle z\rangle=(\sqrt3-1)m_p,\qquad V_{\min}=0,}
$$
a unique <global minimum> of this <Polonyi model>. If $\mu=0$, the <scalar potential> vanishes identically and this uniqueness and <supersymmetry breaking> conclusion do not hold.
For an independent local stability check, the <Hessian matrix> of $U$ at this <global minimum> is diagonal:
$$
U_{aa}=4\sqrt3,\qquad U_{yy}=8-4\sqrt3,\qquad U_{ay}=0.
$$
Since $U$ and its first derivatives vanish there, differentiating the exponential introduces no additional <Hessian matrix> terms at the <global minimum>. For canonically normalized real fluctuations $z=z_0+(s+it)/\sqrt2$, the squared scalar <masses> are therefore
$$
m_s^2=2\sqrt3\,|\mu|^2e^{a_0^2},
\qquad
m_t^2=2(2-\sqrt3)|\mu|^2e^{a_0^2},
$$
which are both positive.
Although the vacuum energy is zero, the <supergravity auxiliary field> is not. In the convention $F^z=-e^{\kappa^2K/2}\overline{D_zW}$, its <vacuum expectation value> obeys
$$
|F^z|=\sqrt3\,|\mu|m_p e^{a_0^2/2}\ne0.
$$
The matter <fermion> transformation contains a term proportional to $F^z\epsilon$, so no nonzero constant <supersymmetry> parameter leaves this vacuum invariant. Hence \b[supersymmetry is spontaneously broken even though the vacuum energy vanishes]: the positive <auxiliary field> contribution cancels the negative gravitational term. The matter <fermion> supplies the <goldstino>, which is absorbed by the <gravitino> in the <super-Higgs mechanism>.
Using the <gravitino mass from a superpotential>, and $z_0+\beta=m_p$, gives
$$
\boxed{m_{3/2}=\kappa^2 e^{\kappa^2K/2}|W|
=|\mu|e^{2-\sqrt3}
=\frac{|F^z|}{\sqrt3m_p}.}
$$
An <auxiliary field> $F^z$ has <mass dimension> two. Defining the <supersymmetry breaking> scale by $\Lambda_{\mathrm{SUSY}}=\sqrt{|F^z|}$, one obtains
$$
\frac{m_{3/2}}{\Lambda_{\mathrm{SUSY}}}
=\frac{\Lambda_{\mathrm{SUSY}}}{\sqrt3m_p}.
$$
Thus the <gravitino> <mass> is small compared with the <supersymmetry breaking> scale when that scale is sub-Planckian, as in the usual hierarchy $|\mu|\ll m_p$. This smallness is gravitational suppression; it is not a conclusion valid for arbitrary choices of the mass parameter.
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