= Solution
Use Planck units and restrict to the physical <Kähler metric> domain
$$
x=S+\bar S>0,\qquad y=T+\bar T-|C|^2>0.
$$
Put $A=ae^{-\alpha S}$ and $w=A+b$, so that $W=C^3+w$. There are no gauged interactions specified, hence no additional <D-term scalar potential>. The <supergravity F-term potential> is $V=e^K(K^{i\bar j}D_iW\overline{D_jW}-3|W|^2)$, with $e^K=1/(xy^3)$. Its <Kähler covariant derivatives of a superpotential> are
$$
D_SW=-\alpha A-\frac W x,\qquad D_TW=-\frac{3W}{y},\qquad D_CW=3C^2+\frac{3\bar C W}{y}.
$$
The nonzero metric entries are
$$
K_{S\bar S}=x^{-2},\quad
\begin{pmatrix}K_{T\bar T}&K_{T\bar C}\\K_{C\bar T}&K_{C\bar C}\end{pmatrix}
=\frac3{y^2}\begin{pmatrix}1&-C\\-\bar C&T+\bar T\end{pmatrix}.
$$
The determinant of this $T,C$ block is $9/y^3>0$, so it is positive definite. With the inverse-index convention $K_{i\bar j}K^{k\bar j}=\delta_i^k$, the raised components are
$$
K^{S\bar S}=x^2,\qquad
\begin{pmatrix}K^{T\bar T}&K^{T\bar C}\\K^{C\bar T}&K^{C\bar C}\end{pmatrix}
=\frac y3\begin{pmatrix}T+\bar T&\bar C\\C&1\end{pmatrix}.
$$
Keeping these complex-conjugate off-diagonal entries in their correct index positions is essential.
To exhibit the <no-scale supergravity> cancellation, write the $T,C$ norm as
$$
K^{i\bar j}D_iW\overline{D_jW}\big|_{T,C}
=\frac y3\left[y|D_TW|^2+|D_CW+\bar C D_TW|^2\right]
=3|W|^2+3y|C|^4.
$$
Here $D_CW+\bar C D_TW=3C^2$. The $3|W|^2$ term cancels the negative term in the <supergravity F-term potential>. The result is \b[the nonnegative scalar potential]
$$
\boxed{V=\frac{|C^3+b+(1+\alpha x)A|^2}{xy^3}+\frac{3|C|^4}{xy^2}.}
$$
This is the <matter-logarithm no-scale potential>, specialized to the exponential <superpotential>.
Choose the <supergravity auxiliary field> convention $F^i=-e^{K/2}K^{i\bar j}\overline{D_jW}$. Substitution into the full inverse <Kähler metric> gives
$$
\boxed{\begin{aligned}
F^S&=e^{K/2}x\,\overline{C^3+b+(1+\alpha x)A},\\
F^T&=e^{K/2}y\,\overline{A+b},\\
F^C&=-e^{K/2}y\,\bar C^2.
\end{aligned}}
$$
An overall auxiliary-field phase is conventional. The $C$ dependence of $F^T$ cancels between the two metric terms. If all three <auxiliary fields> vanished at a finite physical point, $F^C=0$ would force $C=0$, $F^T=0$ would force $A+b=0$, and $F^S=0$ would then force $\alpha x A=0$. Thus $a=b=0$ is necessary. \b[For any nontrivial choice of $a,b$, supersymmetry is broken at every finite physical point.] The allowed exception $a=b=0$, $C=0$ is supersymmetric; the statement cannot hold for all arbitrary parameters without this qualification.
The <no-scale volume runaway criterion> also determines whether the minimum is attained. For fixed $S,C$, differentiation with respect to $y$ gives
$$
\frac{\partial V}{\partial y}=-\frac{3|C^3+b+(1+\alpha x)A|^2}{xy^4}-\frac{6|C|^4}{xy^3}.
$$
It is negative unless both numerators vanish. Therefore a finite stationary vacuum must satisfy
$$
\boxed{C=0,\qquad b+(1+\alpha x)ae^{-\alpha S}=0.}
$$
Every such point is a global minimum with \b[$V_{\min}=0$]. For $a,b\ne0$, setting $r=|b/a|$ and $z=\alpha x>0$ reduces existence to
$$
r=(1+z)e^{-z/2},\qquad \arg a-\alpha\operatorname{Im}S=\arg(-b)\pmod{2\pi}.
$$
The derivative of the magnitude function is $(1-z)e^{-z/2}/2$. It rises from $1$ at the excluded boundary $z=0$ to $2e^{-1/2}$ at $z=1$, then falls to zero. Consequently \b[a finite zero-energy minimum exists exactly when]
$$
\boxed{a,b\ne0,\quad 0<|b/a|\le2e^{-1/2},}
$$
or in the separate trivial family $a=b=0$. There is one positive root for $0<r\le1$, two for $1<r<2e^{-1/2}$, and one double root at equality. If only one of $a,b$ vanishes, or $r$ exceeds this maximum, there is no finite minimum: the infimum is zero as $\operatorname{Re}T\to\infty$. Calling this runaway a finite zero-energy vacuum would be incorrect.
At a nontrivial zero-energy minimum, $W=A+b=-\alpha x A\ne0$, and the auxiliary expectation values reduce to
$$
F^S=F^C=0,\qquad F^T=e^{K/2}y\bar W\ne0.
$$
This explicitly verifies <supersymmetry breaking> with zero <vacuum energy> in <no-scale supergravity>. Both real components of $T$ are exact <flat directions of a scalar potential> on the minimum locus. Away from that locus, $\operatorname{Im}T$ is still absent from the potential, but $\operatorname{Re}T$ is not flat. At a minimum, $C$ is lifted at quartic order: at fixed $S,T$, $V=3|C|^4/[x(T+\bar T)^2]+O(|C|^6)$. A zero quadratic mass does not make $C$ an exact flat direction. The <quartic stabilization at a double no-scale root> likewise gives a vanishing quadratic mass, but no exact flat direction, for $\operatorname{Re}S$ at $z=1$. For $a=b=0$, the whole $C=0$ family has both $S$ and $T$ flat and <supersymmetry> unbroken.
Finally, the <gravitino mass from a superpotential> is
$$
\boxed{m_{3/2}=\frac{|C^3+A+b|}{\sqrt{x}\,y^{3/2}}.}
$$
At a nontrivial zero-energy minimum this becomes
$$
\boxed{m_{3/2}=\frac{\alpha\sqrt{x}|A|}{y^{3/2}}=\frac{\alpha\sqrt{x}|b|}{(1+\alpha x)y^{3/2}}>0.}
$$
It depends on the unfixed volume modulus, so the model does not predict one numerical mass. For $a=b=0$, $C=0$, the <gravitino> is massless.
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