Find three points of order two on elliptic curve. 
Let $C$ be the cubic curve defined by $y^2z = x^3 -xz^2$ where $O = (0:1:0)$ is an inflection point. Find three points of order two in the group $(C, O, +)$.

I know that $2\cdot P = O$ if and only if the line through $O$ and $P$ is tangent to $C$ at $P$, but I'm having trouble finding such points, and I'm not sure if that's the best way to go about it.
 A: You know that $2P=0$ if and only if $l(O,P)$ is tangent to $C$ at $P$. In the affine complement of the line $z=0$, the point $O$ is the point at infinity in the $y$-direction, i.e. vertically. So an affine line passes through $O$ if and only if it is vertical. In this affine complement, the curve $C$ is given by the equation
$$y^2=x^3-x,\tag{1}$$
and implicit differentiation w.r.t. $x$ yields
$$2yy'=3x^2-1,$$
which shows that the slope of the tangent line at $P=(x_0,y_0)\in C$ is given by
$$\frac{3x_0^2-1}{2y_0},$$
whenever $y_0\neq0$ (and $2\neq0$). In particular this shows that the tangent line is not vertical if $y_0\neq0$. So the only candidates for points of order $2$ are points $P=(x_0,y_0)$ with $y_0=0$.
This is a bit roundabout, and the argument becomes clearer if we differentiate $(1)$ w.r.t. $y$, as then
$$2y=3x^2x'-x'=(3x^2-1)x',$$
which shows that the tangent line is vertical if and only if $y=0$ (and $2\neq0$). Solving
$$x^3-x=y^2=0^2=0,$$
indeed yields the solutions $x=0$, $x=-1$ and $x=1$, corresponding to the projective points
$$(0:0:1),\qquad (-1:0:1),\qquad (1:0:1).$$
