Showing that the order of the zero of the $j$ function is $3$ I'm aware that the $j$-invariant can be defined as a function on the upper half-plane $\Pi^{+}$(as seen in the following wikipedia article: https://en.wikipedia.org/wiki/J-invariant):
$$ j(\tau)=1728\frac{(g_{2}(\tau))^{3}}{(g_{2}(\tau)^{3}-27g_{3}(\tau)^{2})} $$
I have to show that the zero of $j(\tau)$ at $p= e^{\frac{2 \pi i}{3}}$ has order 3. Differentiating in a "naive" way gives:
$$ j'(\tau) = \frac{3g_{2}(\tau)^{2}((g_{2}(\tau)^{3}-27g_{3}(\tau)^{2}))- g_{2}(\tau))^{3}(3g_{2}(\tau)^{2})-54g_{3}(\tau))}{(g_{2}(\tau)^{3}-27g_{3}(\tau)^{2})^{2}} $$, this implies that $j'(e^{\frac{2 \pi i}{3}})=0$
In this way, I can show that $$ j''(p) = 0$$, but that $$j'''(p) \neq 0 $$, so the zero of the function is of order 3.
But, I'm not sure if this style of differentiation is correct for a function like the $j$-invariant. Can someone tell me if I'm doing this in an appropriate manner?

$z\to e^{2i\pi /3}z$ is an automorphism of $\Bbb{Z+e^{2i\pi /3}Z}$ so that $g_2(e^{2i\pi 3})=(e^{2i\pi 3})^{-4} g_2(e^{2i\pi 3})\implies g_2(e^{2i\pi 3})=0$.
On the other hand $g_3(e^{2i\pi /3})\ne 0$ and $g_2'(e^{2i\pi /3})\ne 0$ (numerical check).
