= Solution
Because $f$ is a <compactly supported distribution>, the <convolution of distributions with a compactly supported factor> defines $E*f$. The <convolution derivative identity> and part (i) give
$$
\partial_{\bar z}(E*f)=(\partial_{\bar z}E)*f=\delta_0*f=f.
$$
The difference between any two solutions has zero $\partial_{\bar z}$ <distributional derivative>. Part (ii) therefore proves that the general solution of the <partial differential equation> is
$$
\boxed{u=\left(\frac1{\pi z}\right)*f+h,\qquad h\text{ any entire function}.}
$$
This formula is an identity of <distributions>; a pointwise integral is not required for arbitrary $f$.
If $u$ were a <smooth function>, its <Wirtinger derivative> $\partial_{\bar z}u=f$ would also be smooth, contradicting $\operatorname{sing\,supp}f\ne\varnothing$. In fact the <singular support> is exactly preserved here:
$$
\boxed{\operatorname{sing\,supp}u=\operatorname{sing\,supp}f.}
$$
One inclusion follows because a <differential operator> cannot create a singularity where $u$ is smooth. For the reverse inclusion, $E$ is smooth off zero, and its <singular support> is exactly $\{0\}$: if it were smooth near zero, $\partial_{\bar z}E=\delta_0$ would be smooth there. The <support and singular support rules for convolution> give $\operatorname{sing\,supp}(E*f)\subseteq\{0\}+\operatorname{sing\,supp}f$, and adding the smooth <entire function> $h$ leaves this <singular support> unchanged. Thus there is no globally smooth solution when the given <singular support> is nonempty.
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