[proofplan] A point on the line at infinity has homogeneous coordinate $Z = 0$. Substituting this into the defining homogeneous equation of the cubic forces $X^3 = 0$, hence $X = 0$ because $k$ is a field. The remaining coordinate $Y$ must be nonzero, so the projective point $[0:Y:0]$ can be rescaled to $[0:1:0]$. [/proofplan]

[step:Substitute the infinity condition into the cubic equation]Let $P \in C$ be a point lying on the line at infinity. Write \begin{align*} P = [X:Y:Z] \end{align*} with $X,Y,Z \in k$, not all zero. Since $P$ lies on the line at infinity, $Z = 0$. Since $P \in C$, its homogeneous coordinates satisfy \begin{align*} Y^2Z = X^3 + aXZ^2 + bZ^3. \end{align*} Substituting $Z = 0$ gives \begin{align*} 0 = X^3. \end{align*} Because $k$ is a field, it has no nonzero nilpotent elements, so $X = 0$.[/step]

[guided]We begin with an arbitrary point of the cubic on the line at infinity and show that its homogeneous coordinates are forced. Let \begin{align*} P = [X:Y:Z] \end{align*} be a point of $C$ on the line $Z = 0$, where $X,Y,Z \in k$ are homogeneous coordinates and not all of $X,Y,Z$ are zero. The condition that $P$ lies on the line at infinity is exactly the coordinate condition $Z = 0$. The point $P$ also lies on the cubic $C$, so its coordinates satisfy the defining equation \begin{align*} Y^2Z = X^3 + aXZ^2 + bZ^3. \end{align*} Now substitute $Z = 0$ into this equation. The left-hand side becomes $Y^2 \cdot 0 = 0$, and the two terms involving $Z$ on the right-hand side vanish: \begin{align*} 0 = X^3 + aX \cdot 0^2 + b \cdot 0^3 = X^3. \end{align*} Thus $X^3 = 0$. Since $k$ is a field, the product of three copies of $X$ can be zero only when $X = 0$. Therefore every point of $C$ on the line at infinity has the form \begin{align*} P = [0:Y:0]. \end{align*}[/guided]

[step:Use projective nonvanishing and rescale the remaining coordinate] From the previous step, $P = [0:Y:0]$. Since homogeneous coordinates in $\mathbb{P}^2(k)$ cannot all be zero, we must have $Y \neq 0$. Therefore $Y^{-1} \in k$ exists. Rescaling the homogeneous coordinates by the nonzero scalar $Y^{-1}$ gives \begin{align*} [0:Y:0] = [Y^{-1}0:Y^{-1}Y:Y^{-1}0] = [0:1:0]. \end{align*} Hence every point of $C$ on the line at infinity is $[0:1:0]$. Conversely, the point $[0:1:0]$ lies on the line $Z = 0$, and substituting $X = 0$, $Y = 1$, and $Z = 0$ into the defining equation gives \begin{align*} 1^2 \cdot 0 = 0^3 + a \cdot 0 \cdot 0^2 + b \cdot 0^3, \end{align*} so $[0:1:0] \in C$. Therefore \begin{align*} C \cap \{[X:Y:Z] \in \mathbb{P}^2(k) : Z = 0\} = \{[0:1:0]\}. \end{align*} [/step]