[proofplan] We work in the affine chart $Y \ne 0$ containing $O$, with local coordinates $u = X/Y$ and $v = Z/Y$. In these coordinates the curve is cut out near $O$ by a polynomial whose linear term is $v$, so the tangent line at $O$ is $v = 0$, equivalently $Z = 0$ in projective coordinates. Restricting the cubic equation to this tangent line gives $u^3 = 0$, so the local intersection algebra has length $3$. This proves that the tangent line meets the cubic with multiplicity $3$ at $O$; the stated characteristic and nonsingularity hypotheses place the cubic in the usual short Weierstrass setting, although the local computation at $O$ itself uses only this displayed equation near $O$. [/proofplan]

[step:Move to affine coordinates around the point at infinity] Let $U_Y \subset \mathbb{P}^2_k$ denote the standard affine chart where $Y \ne 0$. Define affine coordinates \begin{align*} u &= \frac{X}{Y}, & v &= \frac{Z}{Y}. \end{align*} The point $O = [0:1:0]$ belongs to $U_Y$ and corresponds to $(u,v) = (0,0) \in \mathbb{A}^2_k$. Let \begin{align*} f: \mathbb{A}^2_k \to \mathbb{A}^1_k,\qquad (u,v) \mapsto v - u^3 - auv^2 - bv^3 \end{align*} be the dehomogenized equation of $E$ on $U_Y$. Indeed, substituting $X=uY$ and $Z=vY$ into \begin{align*} Y^2Z - X^3 - aXZ^2 - bZ^3 = 0 \end{align*} and dividing by $Y^3$ gives \begin{align*} v - u^3 - auv^2 - bv^3 = 0. \end{align*} Thus, in the chart $U_Y$, the curve $E$ is locally the affine plane curve $f(u,v)=0$ near $(0,0)$. [/step]

[step:Identify the tangent line from the linear part of the local equation]The polynomial $f$ satisfies \begin{align*} f(0,0) &= 0, & \frac{\partial f}{\partial u}(0,0) &= 0, & \frac{\partial f}{\partial v}(0,0) &= 1. \end{align*} Therefore the degree-one part of $f$ at $(0,0)$ is $v$. Hence the tangent line to $E$ at $O$ in the affine chart $U_Y$ is \begin{align*} v = 0. \end{align*} In homogeneous projective coordinates, the equation $v=Z/Y=0$ on $U_Y$ is the line \begin{align*} Z = 0. \end{align*} So the projective tangent line to $E$ at $O$ is $L = \{Z=0\}$.[/step]

[guided]We now compute the tangent line using only the local equation in the chart containing $O$. The affine equation is \begin{align*} f(u,v)=v-u^3-auv^2-bv^3. \end{align*} At the point corresponding to $O$, namely $(0,0)$, the constant term vanishes: \begin{align*} f(0,0)=0. \end{align*} The tangent line to a nonsingular affine plane curve at a point is cut out by the degree-one part of the local defining equation. We compute that degree-one part by taking the first partial derivatives. The $u$-partial derivative is \begin{align*} \frac{\partial f}{\partial u}(u,v) = -3u^2-av^2. \end{align*} The $v$-partial derivative is \begin{align*} \frac{\partial f}{\partial v}(u,v) = 1-2auv-3bv^2. \end{align*} Evaluating at $(0,0)$ gives \begin{align*} \frac{\partial f}{\partial u}(0,0) &= 0, & \frac{\partial f}{\partial v}(0,0) &= 1. \end{align*} Thus the linear approximation to $f$ at $(0,0)$ is \begin{align*} 0\cdot u + 1\cdot v = v. \end{align*} The affine tangent line is therefore $v=0$. Since $v=Z/Y$ on the chart $Y \ne 0$, this affine equation is the restriction of the projective line $Z=0$. Hence the tangent line at $O$ is $L=\{Z=0\}$.[/guided]

[step:Compute the local intersection algebra with the tangent line] Inside the affine chart $U_Y$, the tangent line $L$ has equation $v=0$. Since $U_Y$ is an affine open neighbourhood of $O$, the [local intersection multiplicity](/page/Local%20Intersection%20Multiplicity) of the projective curves $E$ and $L$ at $O$ is computed in this affine chart. Define $I_O(E,L)$ to be that local intersection multiplicity, namely the length of the local intersection algebra \begin{align*} A = k[u,v]_{(u,v)} \big/ \bigl(f(u,v),v\bigr), \end{align*} where $k[u,v]_{(u,v)}$ is the localization of $k[u,v]$ at the maximal ideal $(u,v)$. Substituting $v=0$ into $f$ gives \begin{align*} f(u,0) = -u^3. \end{align*} Therefore \begin{align*} A &\cong k[u]_{(u)} \big/ (u^3). \end{align*} The quotient $k[u]_{(u)} \big/ (u^3)$ is an Artinian local $k$-algebra because its maximal ideal is generated by the nilpotent class of $u$. As a $k$-[vector space](/page/Vector%20Space), this quotient has basis \begin{align*} 1,\quad u,\quad u^2. \end{align*} Hence its length as a module over itself equals its $k$-dimension, and \begin{align*} \dim_k A = 3. \end{align*} By the definition of $I_O(E,L)$ just fixed above, \begin{align*} I_O(E,L)=\dim_k A=3. \end{align*} [/step]

[step:Conclude that the point at infinity is a flex] The tangent line $L=\{Z=0\}$ intersects $E$ at $O$ with local intersection multiplicity $3$. By the definition of a [flex point](/page/Flex%20Point) on a plane cubic, this means that the tangent line has intersection multiplicity at least $3$ with the cubic at that point. Therefore $O=[0:1:0]$ is a flex point of $E$. [/step]