[proofplan] We compute the tangent line in an affine chart containing $P$, where the curve is defined by the dehomogenized polynomial in two variables. Smoothness says that the affine linear part is nonzero, and the affine tangent line is precisely the zero set of this linear part. We then homogenize that affine tangent equation and use Euler's identity for a homogeneous cubic to identify the coefficient of the chart coordinate with the missing homogeneous partial derivative. Finally, since the first partial derivatives of a cubic are homogeneous of degree $2$, rescaling the representative of $P$ rescales the whole line equation by a nonzero scalar. [/proofplan]

[step:Pass to an affine chart containing the point] At least one of $X_0,Y_0,Z_0$ is nonzero. After permuting the homogeneous coordinates, it suffices to treat the case $Z_0 \neq 0$, because the same argument applies in the charts $X \neq 0$ and $Y \neq 0$ with the variables relabelled. Define $a := X_0/Z_0$ and $b := Y_0/Z_0$. Let \begin{align*} q := (a,b,1) \in k^3 \end{align*} be the normalized representative of $P$ in the affine chart $U_Z := \{[X:Y:Z] \in \mathbb{P}^2_k : Z \neq 0\}$. Define the dehomogenized polynomial $f: k^2 \to k$ by \begin{align*} f(x,y)=F(x,y,1). \end{align*} Then $C_F \cap U_Z$ is identified with the affine plane curve $f(x,y)=0$, and $P$ corresponds to $(a,b) \in k^2$. By the chain rule for this dehomogenization, we have \begin{align*} \frac{\partial f}{\partial x}(a,b)=F_X(q). \end{align*} Similarly, \begin{align*} \frac{\partial f}{\partial y}(a,b)=F_Y(q). \end{align*} [/step]

[step:Write the affine tangent line as the vanishing of the linear part]Since $P$ is smooth on the projective curve, the homogeneous gradient \begin{align*} \left(F_X(q),F_Y(q),F_Z(q)\right) \end{align*} is not the zero vector. In the affine chart $U_Z$, the tangent line to the affine curve $f=0$ at $(a,b)$ is the zero set of the nonzero linear part of $f$ at $(a,b)$: \begin{align*} F_X(q)(x-a)+F_Y(q)(y-b)=0. \end{align*}[/step]

[guided]The purpose of passing to the affine chart is that the tangent line to an affine plane curve has a direct first-order description. In the chart $Z \neq 0$, every point is written uniquely as $[x:y:1]$, and the projective equation $F(X,Y,Z)=0$ becomes the affine equation \begin{align*} f(x,y)=F(x,y,1)=0. \end{align*} The point $P=[X_0:Y_0:Z_0]$ corresponds to $(a,b)=(X_0/Z_0,Y_0/Z_0)$. The first-order Taylor expansion of $f$ at $(a,b)$ has linear part \begin{align*} \frac{\partial f}{\partial x}(a,b)(x-a)+\frac{\partial f}{\partial y}(a,b)(y-b). \end{align*} Because $f(x,y)=F(x,y,1)$, differentiating with respect to the first affine variable gives \begin{align*} \frac{\partial f}{\partial x}(a,b)=F_X(a,b,1)=F_X(q). \end{align*} Differentiating with respect to the second affine variable gives \begin{align*} \frac{\partial f}{\partial y}(a,b)=F_Y(a,b,1)=F_Y(q). \end{align*} Thus the affine tangent line is \begin{align*} F_X(q)(x-a)+F_Y(q)(y-b)=0. \end{align*} Why is this a genuine line? Smoothness of $P$ means that the projective gradient is not zero: \begin{align*} \left(F_X(q),F_Y(q),F_Z(q)\right) \neq (0,0,0). \end{align*} In this affine chart, the tangent direction is controlled by the first-order part of $f$. If $F_X(q)$ and $F_Y(q)$ are not both zero, the displayed equation is visibly a nonzero affine linear equation. If $F_X(q)=F_Y(q)=0$, then smoothness forces $F_Z(q)\neq 0$; Euler's identity below then gives $F_Z(q)=0$ because $q \in C_F$ and $F$ has degree $3$, a contradiction. Hence $F_X(q)$ and $F_Y(q)$ are not both zero in this chart, so the affine tangent equation is nonzero.[/guided]

[step:Homogenize the affine tangent equation] Substitute $x=X/Z$ and $y=Y/Z$ into the affine tangent equation and multiply by $Z$. This gives the projective line in $U_Z$: \begin{align*} F_X(q)X+F_Y(q)Y-\left(aF_X(q)+bF_Y(q)\right)Z=0. \end{align*} Since $F$ is homogeneous of degree $3$, Euler's identity for homogeneous polynomials gives \begin{align*} XF_X+YF_Y+ZF_Z=3F. \end{align*} Evaluating at $q=(a,b,1)$ and using $F(q)=0$ because $P \in C_F$, we obtain \begin{align*} aF_X(q)+bF_Y(q)+F_Z(q)=0. \end{align*} Therefore \begin{align*} -\left(aF_X(q)+bF_Y(q)\right)=F_Z(q), \end{align*} and the homogenized line is \begin{align*} F_X(q)X+F_Y(q)Y+F_Z(q)Z=0. \end{align*} [/step]

[step:Replace the normalized representative by the given representative] The representative $p=(X_0,Y_0,Z_0)$ satisfies $p=Z_0q$. Since each of $F_X,F_Y,F_Z$ is homogeneous of degree $2$, for every $\lambda \in k^\times$ and every $r \in k^3$, we have \begin{align*} F_X(\lambda r)=\lambda^2F_X(r). \end{align*} Also, \begin{align*} F_Y(\lambda r)=\lambda^2F_Y(r). \end{align*} Finally, \begin{align*} F_Z(\lambda r)=\lambda^2F_Z(r). \end{align*} Applying this with $\lambda=Z_0$ and $r=q$, we get \begin{align*} F_X(p)X+F_Y(p)Y+F_Z(p)Z = Z_0^2\left(F_X(q)X+F_Y(q)Y+F_Z(q)Z\right). \end{align*} Because $Z_0^2 \in k^\times$, the two equations define the same projective line. Thus the tangent line at $P$ is \begin{align*} F_X(p)X+F_Y(p)Y+F_Z(p)Z=0. \end{align*} The same homogeneity calculation also shows that replacing $p$ by any $\lambda p$ with $\lambda \in k^\times$ multiplies the displayed equation by $\lambda^2$, so the projective line is independent of the chosen representative. [/step]