[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 $L = \{Z=0\}$.
[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]
[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]
