= Solution
The exponent in the original PDF is $e^{-\alpha S}$ with real $\alpha>0$, not the $e^{-aS}$ produced by the local TeX. Use that authoritative expression. Put
$$
s=S+\bar S>0,\qquad t=T+\bar T,\qquad q=t-|C|^2>0,\qquad w(S)=ae^{-\alpha S}+b,\qquad W=C^3+w(S).
$$
The domain conditions ensure a real <Kähler potential> and positive <Kähler metric>. There are no gauge multiplets specified, so the <scalar potential> is the <supergravity F-term potential>
$$
V=e^K\left(K^{i\bar j}D_iW\overline{D_jW}-3|W|^2\right),\qquad D_iW=W_i+K_iW,\qquad e^K=\frac1{s q^3}.
$$
The <Kähler covariant derivative of a superpotential> and metric entries are
$$
D_SW=w_S-\frac Ws,\qquad D_TW=-\frac{3W}{q},\qquad D_CW=3C^2+\frac{3\bar C W}{q},\qquad K_{S\bar S}=s^{-2},
$$
$$
(K_{i\bar j})_{i,j=T,C}=\frac3{q^2}\begin{pmatrix}1&-C\\-\bar C&t\end{pmatrix}.
$$
State the inverse with its indices explicitly, since transposing the off-diagonal complex entries would change the answer:
$$
K^{S\bar S}=s^2,\qquad K^{T\bar T}=\frac{qt}{3},\qquad K^{T\bar C}=\frac{q\bar C}{3},\qquad K^{C\bar T}=\frac{qC}{3},\qquad K^{C\bar C}=\frac q3.
$$
Here $\sum_jK_{i\bar j}K^{k\bar j}=\delta_i^k$, so the displayed upper-index array is the transpose of the ordinary <matrix> inverse of the displayed lower-index array.
Substitution shows that the cross terms involving $W$ cancel and that the $T,C$ sector gives
$$
K^{i\bar j}D_iW\overline{D_jW}=3|W|^2+\frac q3|W_C|^2\quad(i,j=T,C).
$$
The first term cancels the universal negative term. Thus the <no-scale supergravity> potential is
$$
\boxed{V=\frac{|s w_S-W|^2+3q|C|^4}{s q^3}=\frac{|C^3+b+(1+\alpha s)ae^{-\alpha S}|^2+3q|C|^4}{s q^3}\ge0.}
$$
This positivity is an algebraic cancellation, not an assumption that the individual Kähler derivatives vanish.
Choose the <supergravity auxiliary field> convention $F^i=-e^{K/2}K^{i\bar j}\overline{D_jW}$. Direct multiplication gives
$$
\boxed{F^S=e^{K/2}s(\bar W-s\bar w_S),\qquad F^T=e^{K/2}q\bar w,\qquad F^C=-e^{K/2}q\bar C^2.}
$$
A conventional common phase or overall sign on the <auxiliary fields> changes none of the vanishing conditions. A supersymmetric configuration requires every $F^i$ to vanish. First $F^C=0$ implies $C=0$, then $F^T=0$ implies $w=0$, and $F^S=0$ implies $w_S=0$. Since $w_S=-\alpha a e^{-\alpha S}$, at a finite point of the physical domain this is possible only when $a=0$, followed by $b=0$. Consequently \b[If $(a,b)\ne(0,0)$, every finite configuration has a nonzero <auxiliary field>, so any finite vacuum breaks <supersymmetry>]. The printed request cannot hold for arbitrary parameters without this exception: $a=b=0$, $C=0$ is a zero-energy supersymmetric family with both $S$ and $T$ unfixed.
With the stipulated vanishing <vacuum expectation value> of $C$, define $A(S)=b+(1+\alpha s)ae^{-\alpha S}$. The potential reduces to $|A|^2/(s t^3)$. Its derivative along $t$ is $-3V/t$, so a finite stationary point must have $A=0$. Such a point is a global minimum of the full nonnegative potential, since both squares vanish at $C=0$. The <finite zero-energy vacuum for a single exponential superpotential> therefore satisfies
$$
\boxed{C=0,\qquad b=-(1+\alpha s)ae^{-\alpha S},\qquad V=0.}
$$
For nonzero $a,b$, write $S=\sigma+i\chi$ and $x=\alpha\sigma>0$. The modulus and phase conditions are
$$
\rho\equiv\left|\frac ba\right|=(1+2x)e^{-x},\qquad e^{-i\alpha\chi}=-\frac{b/a}{\rho}.
$$
The function $(1+2x)e^{-x}$ increases up to $x=1/2$ and then decreases to zero; its maximum is $2e^{-1/2}$. Hence a finite minimum exists exactly when
$$
\boxed{a b\ne0,\qquad0<|b/a|\le2e^{-1/2}.}
$$
There is one positive solution for $0<\rho\le1$, two for $1<\rho<2e^{-1/2}$, and one coalesced solution at the upper bound. The formal $x=0$ solution at $\rho=1$ is outside $s>0$. At each allowed $x$, the phase fixes $\chi$ modulo $2\pi/\alpha$. Equivalently $x=-\tfrac12-W_k(-\rho/(2\sqrt e))$, using the real branches of the <Lambert W function> and retaining only $x>0$.
At any of these nontrivial minima, $w=-\alpha s ae^{-\alpha S}\ne0$. The <auxiliary fields> become
$$
\boxed{F^S=F^C=0,\qquad F^T=e^{K/2}t\bar w\ne0.}
$$
This is a <supersymmetry breaking> Minkowski minimum in which the nonzero <auxiliary field> belongs to $T$. Both real components of $T$ are exact <flat directions of a scalar potential> at the minimum. They change the metric and auxiliary-field magnitudes but not the zero potential. Generically the two real components of $S$ are fixed. At the coalesced solution $x=1/2$, the radial quadratic restoring term vanishes, but the leading restoring term is quartic; this is not an additional exact <flat direction of a scalar potential>. The matter field $C$ likewise has a positive leading quartic potential, not an exact <flat direction of a scalar potential>, despite its zero quadratic mass here.
The exceptional parameter cases must also be stated. If $a=0,b\ne0$ or $a\ne0,b=0$, no finite zero-energy minimum with $C=0$ exists. The same holds when $|b/a|>2e^{-1/2}$. Because the $t$ derivative is nonzero at any positive-energy point on $C=0$, none is a finite minimum there; the energy approaches zero along the runaway $t\to\infty$. If $a=b=0$, all physical $S,T$ at $C=0$ give supersymmetric zero-energy minima. Thus an unconditional finite minimum or unconditional breaking would be a false claim for the printed arbitrary parameters.
For the homogeneous model, assume $\Gamma>0$ is twice differentiable, of degree one in the real moduli $\tau_i$, and that its <Hessian matrix> for $K$ is invertible on the sector considered. Differentiate $K=-3\log\Gamma$:
$$
K_i=-\frac{3\Gamma_i}{\Gamma},\qquad K_{ij}=\frac{3\Gamma_i\Gamma_j}{\Gamma^2}-\frac{3\Gamma_{ij}}{\Gamma}.
$$
The two <Euler theorem for homogeneous functions> identities in the question imply
$$
\boxed{\tau_iK_{ij}=\frac{3\Gamma_j}{\Gamma}.}
$$
Multiplying by the inverse gives $K^{-1}_{ij}\Gamma_j/\Gamma=\tau_i/3$, so contracting once more yields
$$
\boxed{\frac{\Gamma_iK^{-1}_{ij}\Gamma_j}{\Gamma^2}=\frac13,\qquad K_iK^{-1}_{ij}K_j=3.}
$$
This is the <no-scale identity from degree-one homogeneity>. It does not assert that every degree-one function gives a positive or invertible metric: for example $\Gamma=\tau_1+\tau_2$ has a rank-one Kähler Hessian, and its inverse is undefined. The inverse hypothesis is necessary.
For complex moduli with $\tau_i=T_i+\bar T_i$, the same derivatives are the mixed <Kähler metric>; choosing real parts with a factor of two only introduces factors that cancel in the contraction. If $W$ is independent of these moduli, $D_iW=K_iW$, and their contribution is $3e^K|W|^2$. It cancels the universal $-3e^K|W|^2$, leaving no <scalar potential> from this isolated no-scale sector. \b[Other chiral sectors can still contribute positive terms]. For this conclusion in a larger theory, the displayed metric must be the appropriate decoupled no-scale block, or the full inverse metric must itself satisfy the corresponding identity; arbitrary mixed additions to $K$ do not inherit the cancellation automatically.
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