= Solution
All derivatives below are <left Grassmann derivatives>. Decompose the <supersymmetric covariant derivatives> into odd pieces
$$
D_\alpha=\partial_\alpha+A_\alpha,\qquad
\bar D_{\dot\beta}=\bar\partial_{\dot\beta}+B_{\dot\beta},
\qquad
A_\alpha=i\sigma^\mu_{\alpha\dot\gamma}\bar\theta^{\dot\gamma}\partial_\mu,
\quad B_{\dot\beta}=i\theta^\gamma\sigma^\mu_{\gamma\dot\beta}\partial_\mu.
$$
When these operators act on an arbitrary <superfield>, the <graded Leibniz rule> gives
$$
\{\partial_\alpha,B_{\dot\beta}\}=i\sigma^\mu_{\alpha\dot\beta}\partial_\mu,
\qquad
\{A_\alpha,\bar\partial_{\dot\beta}\}=i\sigma^\mu_{\alpha\dot\beta}\partial_\mu.
$$
The <anticommutator> of the two pure Grassmann derivatives vanishes. Also $\{A_\alpha,B_{\dot\beta}\}=0$: the spacetime derivatives commute, while the $\theta$ and $\bar\theta$ coefficients anticommute. Adding these terms proves the <supercovariant derivative algebra with left derivatives>,
$$
\boxed{\{D_\alpha,\bar D_{\dot\beta}\}=2i\sigma^\mu_{\alpha\dot\beta}\partial_\mu}.
$$
For two unbarred derivatives, neither $\partial_\alpha$ differentiates the $\bar\theta$ coefficient of the other operator. Their coefficient products anticommute, so $\boxed{\{D_\alpha,D_\beta\}=0}$. The same argument with barred and unbarred coordinates exchanged gives $\boxed{\{\bar D_{\dot\alpha},\bar D_{\dot\beta}\}=0}$. These signs follow the derivatives printed in this paper; a convention replacing $\bar D$ by $-\bar D$ reverses the mixed sign as well.
Set $y^\mu=x^\mu+i\theta^\alpha\sigma^\mu_{\alpha\dot\beta}\bar\theta^{\dot\beta}$. Left differentiation gives
$$
\bar\partial_{\dot\alpha}(\theta\sigma^\mu\bar\theta)
=-\theta^\beta\sigma^\mu_{\beta\dot\alpha},
\qquad\bar D_{\dot\alpha}y^\mu=0.
$$
The <chain rule> therefore makes $\bar D_{\dot\alpha}$ simply $\bar\partial_{\dot\alpha}$ at fixed $y,\theta$. The <chiral superfield> constraint $\bar D_{\dot\alpha}\Phi=0$ means that $\Phi$ is independent of $\bar\theta$ in these coordinates. Since there are only two independent entries of $\theta$, its <Grassmann algebra> expansion terminates at degree two. Naming the coefficients gives the <chiral-superfield component expansion>
$$
\boxed{\Phi(y,\theta)=\phi(y)+\sqrt2\theta\psi(y)+\theta\theta F(y)}.
$$
Here $\phi$ is a <complex scalar field>, $\psi$ a <Weyl spinor>, and $F$ a nonpropagating <auxiliary field>; the $\sqrt2$ is their conventional normalization.
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