= Solution
Assume global <four-dimensional N=1 supersymmetry> with canonical <kinetic terms> and no explicit <gaugino> mass. Write $W_i=\partial_iW$ and $W_{ij}=\partial_i\partial_jW$. To match the displayed stationarity equation, use $F^i=\overline{W_i}$; this differs by a minus sign from the eliminated <auxiliary field> in the component convention of Solution 2. The real <D-terms> are $D^a=-g^a(\bar\phi_i(T^a)^i{}_j\phi^j+\xi^a)$, with Hermitian <gauge generators>. A constant <Fayet–Iliopoulos term> is permitted only for an appropriate Abelian factor. The <scalar potential> is
$$
V=W_i\overline{W_i}+\frac12\sum_a(D^a)^2.
$$
Differentiate with respect to $\phi^i$, holding $\bar\phi$ independent:
$$
\partial_iD^a=-g^a\bar\phi_j(T^a)^j{}_i,\qquad \partial_iV=W_{ij}F^j-\sum_ag^aD^a\bar\phi_j(T^a)^j{}_i.
$$
Thus <stationary point> conditions gives the first requested identity. <Gauge invariance> of the <superpotential> gives $W_i(T^a)^i{}_j\phi^j=0$. Conjugating it and using Hermiticity gives $\bar\phi_j(T^a)^j{}_iF^i=0$, the second condition in the form needed below.
At the vacuum, define $H_{ij}=\langle W_{ij}\rangle$ and $G_{ia}=g^a\langle\bar\phi_j\rangle(T^a)^j{}_i$. The two conditions combine as
$$
\boxed{M_0\begin{pmatrix}\langle F\rangle\\\langle D\rangle\end{pmatrix}=0,\qquad M_0=\begin{pmatrix}H&-G\\-G^T&0\end{pmatrix}.}
$$
Here the upper equation is stationarity and the lower equation is conjugated <gauge invariance>. The <chiral-superfield fermion mass matrix> contributes $-\frac12H_{ij}\psi^i\psi^j$, and the gauge <Yukawa interaction> contributes the mixed chiral-fermion/<gaugino> bilinear. With canonical <Weyl fermion> and <gaugino> <kinetic terms>, and choosing the <gaugino> phase to put the mixed bilinear in real-sign convention, these terms are
$$
\mathcal L_{\mathrm{mass}}=-\frac12\begin{pmatrix}\psi&\lambda\end{pmatrix}\mathcal M\begin{pmatrix}\psi\\\lambda\end{pmatrix}+\mathrm{h.c.},\qquad \mathcal M=\begin{pmatrix}H&-\sqrt2G\\-\sqrt2G^T&0\end{pmatrix}.
$$
The factor $\sqrt2$ is important: the <matrix> directly acting on the printed $(F,D)$ is the same mass bilinear with rescaled <gauginos>, rather than literally the canonical <matrix>. In detail, for $S=\operatorname{diag}(I,\sqrt2I)$,
$$
\mathcal M=SM_0S,\qquad \boxed{\mathcal M\begin{pmatrix}\langle F\rangle\\\langle D\rangle/\sqrt2\end{pmatrix}=0.}
$$
Since <supersymmetry breaking> makes this vector nonzero, there is at least one zero mass <eigenvalue>. A normalized massless <fermion> is
$$
\eta=\frac{\sum_i\langle F^i\rangle^*\psi^i+\sum_a\langle D^a\rangle\lambda^a/\sqrt2}{\left(\sum_i|\langle F^i\rangle|^2+\frac12\sum_a\langle D^a\rangle^2\right)^{1/2}}.
$$
With the consistent <gaugino> phase, the inhomogeneous vacuum <supersymmetry transformation> of $(\psi,\lambda)$ is proportional to the same vector $(F,D/\sqrt2)$. Therefore $\eta$ shifts by a nonzero constant times the transformation parameter: it is the <Goldstino>, the <Goldstone fermion> of broken global <supersymmetry>. The argument guarantees \b[at least one <Goldstino> zero mode], not the absence of other massless fermions.
For the second calculation, first take only canonical <chiral superfields> and <F-term> breaking. At a stationary vacuum $v$, let $h=\phi-v$, $m_{ij}=W_{ij}(v)$ and
$$
A_{ij}=\sum_k\overline{W_{ki}}W_{kj},\qquad B_{ij}=\sum_k\overline{W_k}W_{kij}.
$$
These follow by differentiating $V=\sum_k|W_k|^2$ twice. The quadratic <scalar potential> is
$$
V^{(2)}=h^\dagger Ah+\frac12(h^TBh+\mathrm{h.c.}),\qquad A=m^\dagger m.
$$
The doubled scalar <mass matrix> has blocks
$$
M_s^2=\begin{pmatrix}A&B^\dagger\\B&A^T\end{pmatrix}.
$$
Transforming to the $2N$ canonically normalized real components of $h$ preserves its <trace>, so the sum of real scalar mass squares is $\operatorname{tr}M_s^2=2\operatorname{tr}A$. The <Weyl fermion> mass squares are the <eigenvalues> of $m^\dagger m=A$ and their sum is $\operatorname{tr}A$. Each Weyl mass eigenstate has two physical spin states. The printed <supertrace> has scalar sign $-1$ and fermion weight $+2$, hence
$$
\boxed{\operatorname{STr}M^2=-\operatorname{tr}M_s^2+2\operatorname{tr}(m^\dagger m)=-2\operatorname{tr}A+2\operatorname{tr}A=0.}
$$
The frequently used boson-minus-fermion definition has the opposite overall sign and gives the same zero. The mixing block $B$ can split scalar masses but cannot change this <trace>. This proves the <tree-level supertrace mass sum rule>; it uses canonical <kinetic terms> and a renormalizable global theory, rather than a general <supergravity> or noncanonical effective theory.
In the <MSSM>, trying to make all <squarks> and <sleptons> heavy through direct canonical tree-level <F-term> breaking encounters this constraint. In an unbroken electric-charge and color sector, the scalar mass-squared <trace> is twice the <trace> for its ordinary fermionic partners. More explicitly, take a unit vector $u$ minimizing $u^\dagger Au$. For $h=e^{i\alpha}u$, choose $\alpha$ so that $\operatorname{Re}(e^{2i\alpha}u^TBu)\leq0$. The curvature along this real scalar direction is then at most $u^\dagger Au$, so at least one scalar mass square is no larger than the smallest fermion mass square in that sector. For one isolated pair the scalar masses are $m_f^2\pm|B|$. Thus light <quarks> can force an unacceptably light <squark>, or a tachyon, under these assumptions. A <hidden supersymmetry-breaking sector> with mediation produces effective <soft supersymmetry breaking>; radiative effects, noncanonical <kinetic terms>, or <supergravity> can change the assumptions and allow a viable spectrum. The <trace> over an arbitrary enlarged theory alone would not locate the light state in the visible sector; the MSSM conclusion uses the conserved-charge block structure.
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