I am trying to prove that the order of the two points $P_{\pm}=(0,\pm\sqrt{-g_3})$ is three on the elliptic curve $y^2=4x^3-g_3$, for $g_3 \not= 0$, defined over $\mathbb{P}^2_{\mathbb{C}}$. Here's an attempt:
To find $P_+ + P_+$ I differentiate the equation of the curve and get the slope $m=\dfrac{12x_1^2}{2\sqrt{-g_3}}$, which gives slope of the tangent $0$ at $P_+$, so the tangent is the line $y=\sqrt{-g_3}$. Plugging into the equation of the curve I get $x=0$. It follows that the point $(0,\sqrt{-g_3})$, which is equal to $P_+$, is also equal to $-2P_+$, which means that $P_+$ has order 3.
Is this the right way to do it?
The slope of the tangent line to $E:y^2=4x^3-g$ at $P=(x_1,y_1)$ is $$m = \frac{6x_1^2}{y_1}$$ as long as $y_1\neq 0$ (i.e., as long as $P$ is not a point of order $2$). Now, let $P=(0,\sqrt{-g})$. Then, the tangent line $L$ to $P$ is a line of slope $0$ passing through $P$, i.e., $y=\sqrt{-g}$. Thus, $L$ intersects $E$ at three points, which can be found by plugging $y=\sqrt{-g}$ in the equation for $E$ and this implies $x=0$. Hence, there is only one point in the intersection, namely $P$ (which means $P$ is a point of triple intersection, and not just double intersection point as it is to be expected since $L$ is tangent to $E$ at $P$). In other words, $P$ is an inflection point for $E$ and the tangent line at $P$ has a triple point of intersection with $E$. Thus, $P+P+P=\mathcal{O}$, or equivalently $-P=2P$. Therefore, $3P=\mathcal{O}$.