= Solution
The <infinite product> converges locally uniformly in $\mathbb H$, because $\sum_{n\geq1}|q|^n<\infty$ uniformly on compact subsets. None of its factors vanishes there, and the convergent logarithm of its tail shows that the product is nonzero. Logarithmic differentiation, justified by locally <uniform convergence>, gives
$$
\Theta\log F=1-24\sum_{n\geq1}\frac{nq^n}{1-q^n}
=1-24\sum_{m\geq1}\sigma_1(m)q^m=E_2.
$$
The middle equality groups the absolutely convergent terms $nq^{nr}$ by $m=nr$. Thus $F'/F=2\pi iE_2$.
Let $H(\tau)=\tau^{-12}F(-1/\tau)/F(\tau)$. It is nonzero and <holomorphic> on the simply connected upper half-plane. Using the printed transformation $E_2(-1/\tau)=\tau^2E_2(\tau)+6\tau/(\pi i)$, its logarithmic <derivative> is
$$
\frac{H'}H=-\frac{12}{\tau}+
\frac{2\pi i}{\tau^2}E_2(-1/\tau)-2\pi iE_2(\tau)=0.
$$
So $H$ is constant. At $\tau=i$ we have $-1/i=i$ and $i^{-12}=1$, giving $H(i)=1$. Therefore $F(-1/\tau)=\tau^{12}F(\tau)$. Also $F(\tau+1)=F(\tau)$ directly from $q=e^{2\pi i\tau}$. Since these two transformations generate $SL_2(\mathbb Z)$, $F$ has the weight-twelve <modular form> transformation law.
At the unique <modular cusp> its expansion is $F=q+O(q^2)$, so it extends holomorphically with zero value. Hence
$$
\boxed{F\in S_{12}(SL_2(\mathbb Z)),\qquad \Theta F=E_2F.}
$$
This is the <modular discriminant>. The transformed argument contributes $6\tau/(\pi i)$; the converted TeX's numerator $67$ is a transcription error, not the printed relation.
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