= Solution
Use $\Theta=(2\pi i)^{-1}d/d\tau$. For $\gamma=\begin{pmatrix}a&b\\c&d\end{pmatrix}$, differentiating the weight-$k$ law gives
$$
(\Theta f)(\gamma\tau)
=(c\tau+d)^{k+2}\Theta f(\tau)
+\frac{kc}{2\pi i}(c\tau+d)^{k+1}f(\tau).
$$
The transformations of $E_2$ under the generators $S,T$, and the chain rule for their compositions, give
$$
E_2(\gamma\tau)=(c\tau+d)^2E_2(\tau)+\frac{12c}{2\pi i}(c\tau+d).
$$
Equivalently, one can check just $S,T$ in the next cancellation, since they generate the group. Substituting these formulas into
$$
D_kf=\Theta f-\frac{k}{12}E_2f
$$
cancels the extra term and gives $(D_kf)(\gamma\tau)=(c\tau+d)^{k+2}D_kf(\tau)$. Both constituents are <holomorphic> on $\mathbb H$ and have <Fourier expansions of a modular form> with nonnegative exponents, so $D_kf$ is <holomorphic> at the <modular cusp>. Thus
$$
\boxed{D_k:M_k\longrightarrow M_{k+2}.}
$$
This operator is the <Serre derivative>.
If $f=\sum_{n\geq0}a_nq^n$, then its <derivative> has constant coefficient zero, while $E_2$ has constant coefficient one. Therefore
$$
a_0(D_kf)=-\frac{k}{12}a_0(f).
$$
For $k\ne0$ this vanishes exactly when $a_0(f)=0$, proving the claimed <cusp form> equivalence in the intended positive-weight setting. At \b[weight zero the printed equivalence is false]: $f=1$ is a noncuspidal weight-zero <modular form>, but $D_0f=0$ belongs to $S_2$. Indeed all weight-zero forms are constants by <compactness> of $X(1)$. Negative-weight level-one <modular forms> are zero by the <valence formula for the modular group>, so there is no further nonzero exception. The transformation and holomorphy conclusion for $D_kf$ remains true at weight zero.
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