= Solution
For arbitrary <linear operators>, expansion of the <anticommutators> gives
$$
\{X,Y\}Z-Y\{X,Z\}=(XY+YX)Z-Y(XZ+ZX)=XYZ-YZX=[X,YZ].
$$
Now use the <commutator product rule> $[VX,W]=V[X,W]+[V,W]X$, with $W=YZ$, twice. The result is
$$
\boxed{[VX,YZ]=V\{X,Y\}Z-VY\{X,Z\}+\{V,Y\}ZX-Y\{V,Z\}X.}
$$
No grading assumption is needed for these algebraic identities.
For the <Dirac field>, supplement the supplied equal-time <canonical anticommutation relations> with $\{\psi_a(x),\psi_b(y)\}=\{\psi_a^\dagger(x),\psi_b^\dagger(y)\}=0$. Define the <fermion number> density $n(\mathbf x)=\sum_a\psi_a^\dagger(\mathbf x)\psi_a(\mathbf x)$. In the preceding identity take $V=\psi_a^\dagger(\mathbf x)$, $X=\psi_a(\mathbf x)$, $Y=\psi_b^\dagger(\mathbf y)$, $Z=\psi_c(\mathbf z)$ and sum over $a$. Only its first and last <anticommutators> survive, giving
$$
\boxed{[n(\mathbf x),\psi_b^\dagger(\mathbf y)\psi_c(\mathbf z)]
=\left[\delta^{(3)}(\mathbf x-\mathbf y)-\delta^{(3)}(\mathbf x-\mathbf z)\right]
\psi_b^\dagger(\mathbf y)\psi_c(\mathbf z).}
$$
This is an identity of operator-valued <distributions>. Point splitting, or smearing with test functions, gives a precise meaning to coincident bilinears. <Normal ordering> only subtracts a constant from the density and does not alter this <commutator>.
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