[proofplan] We restrict the defining homogeneous equation of $C$ to the line $L_\infty$, where $Z=0$. The equation then reduces to $X^3=0$, forcing the first homogeneous coordinate to vanish. A projective point with both $X=0$ and $Z=0$ must have nonzero middle coordinate, and projective scaling identifies all such points with $[0:1:0]$. [/proofplan] [step:Restrict the cubic equation to the line at infinity] Let $P \in C(k) \cap L_\infty(k)$ be a $k$-rational projective point. Choose homogeneous coordinates over $k$ for $P$ and write \begin{align*} P = [X_0:Y_0:Z_0] \end{align*} with $(X_0,Y_0,Z_0) \in k^3 \setminus \{(0,0,0)\}$. Since $P \in L_\infty$, we have $Z_0 = 0$. Since $P \in C$, its coordinates satisfy \begin{align*} Y_0^2 Z_0 = X_0^3 + aX_0Z_0^2 + bZ_0^3. \end{align*} Substituting $Z_0 = 0$ gives \begin{align*} 0 = X_0^3. \end{align*} Because $k$ is a field, $k$ has no nonzero nilpotent elements, so $X_0 = 0$. [guided] Let $P \in C(k) \cap L_\infty(k)$ be any $k$-rational point. To prove that the $k$-rational intersection contains at most one point, we show that the homogeneous coordinates of $P$ over $k$ are forced into one projective equivalence class. Choose a representative \begin{align*} P = [X_0:Y_0:Z_0] \end{align*} where $(X_0,Y_0,Z_0) \in k^3 \setminus \{(0,0,0)\}$. This nonzero condition is part of the definition of a projective point: the zero triple does not define a point of $\mathbb{P}^2_k$. The condition $P \in L_\infty$ means exactly that the defining linear equation of $L_\infty$ vanishes at $P$, hence \begin{align*} Z_0 = 0. \end{align*} The condition $P \in C$ means that the defining homogeneous cubic equation of $C$ vanishes at $P$, so \begin{align*} Y_0^2 Z_0 = X_0^3 + aX_0Z_0^2 + bZ_0^3. \end{align*} Substituting $Z_0 = 0$ eliminates every term containing $Z_0$ and gives \begin{align*} 0 = X_0^3. \end{align*} Since $k$ is a field, there are no nonzero nilpotent elements. Therefore the equality $X_0^3=0$ forces \begin{align*} X_0 = 0. \end{align*} Thus every point of $C \cap L_\infty$ has a representative of the form $[0:Y_0:0]$. [/guided] [/step] [step:Normalize the remaining projective coordinate] From the previous step, every $P \in C \cap L_\infty$ has a representative \begin{align*} P = [0:Y_0:0]. \end{align*} Since $(0,Y_0,0) \ne (0,0,0)$, we have $Y_0 \ne 0$. Let $k^\times := k \setminus \{0\}$ denote the multiplicative group of nonzero elements of $k$. Therefore $Y_0 \in k^\times$, and projective scaling by $Y_0^{-1}$ gives \begin{align*} [0:Y_0:0] = [0:1:0]. \end{align*} Hence \begin{align*} C \cap L_\infty \subset \{[0:1:0]\}. \end{align*} [/step] [step:Verify that the normalized point lies on the cubic and on the line at infinity] The point $O := [0:1:0] \in \mathbb{P}^2_k$ is well-defined because $(0,1,0) \ne (0,0,0)$. Its third homogeneous coordinate is $0$, so $O \in L_\infty$. Evaluating the defining equation of $C$ at $(X,Y,Z)=(0,1,0)$ gives \begin{align*} 1^2 \cdot 0 = 0^3 + a \cdot 0 \cdot 0^2 + b \cdot 0^3, \end{align*} so both sides are equal to $0$. Therefore $O \in C$, and hence \begin{align*} \{[0:1:0]\} \subset C \cap L_\infty. \end{align*} Combining this inclusion with the reverse inclusion from the previous step yields \begin{align*} C(k) \cap L_\infty(k) = \{[0:1:0]\}. \end{align*} [/step]