Let $\Delta(z)=(2\pi)^{12} q\prod_{n=1}^{\infty}(1-q^n)^{24}$ with $q=e^{2\pi i z}$ be the modular discriminant.
I found $$\left(\dfrac{\Delta(6z)\Delta(z)}{\Delta(3z)\Delta(2z)}\right)^{1/2}=q\prod_{n=0}^{\infty}(1-q^{6n+1})^{12}(1-q^{6n+5})^{-12}$$ and I don't understand why.
I get $$q\prod_{n=1}^{\infty}((1-q^{6n})(1-q))^{12}((1-q^{3n})(1-q^{2n}))^{-12}.$$ Is that the same?
Thanks.
Note that $q^{n}$ corresponds to $nz$. Hence we have \begin{align} F(q) &= \left(\frac{\Delta(6z)\Delta(z)}{\Delta(3z)\Delta(2z)}\right)^{1/2}\notag\\ &= \left(\frac{q^{6}\prod_{n = 1}^{\infty}(1 - q^{6n})^{24}\cdot q\prod_{n = 1}^{\infty}(1 - q^{n})^{24}}{q^{3}\prod_{n = 1}^{\infty}(1 - q^{3n})^{24}\cdot q^{2}\prod_{n = 1}^{\infty}(1 - q^{2n})^{24}}\right)^{1/2}\notag\\ &= q\prod_{n = 1}^{\infty}\left(\frac{(1 - q^{n})(1 - q^{6n})}{(1 - q^{2n})(1 - q^{3n})}\right)^{12}\text{ (you have reached till this point)}\notag\\ &= q\prod_{n = 1}^{\infty}\left(\frac{(1 - q^{n})(1 - q^{6n})}{(1 - q^{2n})(1 - q^{3n})}\right)^{12}\notag\\ &= q\prod_{n = 1}^{\infty}\left(\frac{1 - q^{2n - 1}}{1 - q^{3(2n - 1)}}\right)^{12}\notag\\ &= q\prod_{n = 1}^{\infty}\left(\frac{1 - q^{2n - 1}}{1 - q^{6n - 3}}\right)^{12}\notag\\ &= q\prod_{n = 1}^{\infty}\left(\frac{(1 - q^{6n - 1})(1 - q^{6n - 3})(1 - q^{6n - 5})}{1 - q^{6n - 3}}\right)^{12}\notag\\ &= q\prod_{n = 1}^{\infty}\{(1 - q^{6n - 1})(1 - q^{6n - 5})\}^{12}\notag\\ &= q\prod_{n = 0}^{\infty}\{(1 - q^{6n + 1})(1 - q^{6n + 5})\}^{12}\notag\\ &= q(1 - q)^{12}(1 - q^{5})^{12}(1 - q^{7})^{12}(1 - q^{11})^{12}\dots\notag\\ &= q(1 - 12q + 66q^{2} - 220q^{3} + 495q^{4} - 792q^{5} + \cdots)(1 - 12q^{5} + \cdots)\cdots\notag\\ &= q(1 - 12q + 66q^{2} - 220q^{3} + 495q^{4} - 804q^{5} + \cdots)\notag\\ &= q - 12q^{2} + 66q^{3} - 220q^{4} + 495q^{5} - 804q^{6} + \cdots\notag \end{align} So the last exponent in your result should be $12$ instead of $-12$. Luckily all of the steps require only basic algebraic manipulation and there is no need to invoke any non-trivial result from the theory of modular forms.