Finding the equation of a tangent line to a projective curve at a non-singular point.

I am currently working through Fulton's Algebraic Curves and I have attempted the following problem: $$\text{Let P be a simple (non-singular) point on }F\text{ . Show that the tangent line to }F \text{ at } P\text{ has the equation }F_X (P )X + F_Y (P )Y + F_Z (P )Z = 0$$

My solution thus far goes as follows:

Let $F$ be of degree $d$. Without loss of generality, let $P\in U_3$ and let $P = [a : b : 1]$, we know that $m_P(F) = m_{P_*}(F_*) = 1$, where $P_*=(a,b)$. Then let $G = F_*\left(x + a, y + b\right)$. Since the multiplicity at $P$ is $1$, if we write $G$ as a sum of homogeneous polynomials of ascending degree $G = G_1 + \dots + G_m$, $G_1$ will have degree $1$ Hence, the tangent to $F_*$ at $P_*$ will be of the form $G_1 = u x + v y$. Since this is a tangent to $F_*$ at $P_*$, we also have that $ua+vb=0$.

Moreover, by Euler's equation, we get that $dF = x F_x + y F_y + z F_z$. Moreover, $dF(P) = 0 \Rightarrow a F_x(P) + b F_y(P) + F_z(P)=0$

Now computing the derivatives $F_x(P) = \left(\frac{d}{dx} F_*\right)\left(a, b\right) = u$, $F_y(P) = \left(\frac{d}{dy} F_*\right)\left(a, b\right) = v$. This forces $F_z(P)=0$, as $a u + b v=0$.

Thus, it remains to show that the intersection multiplicity of $ax+by$ with $F$ at $P$ is bigger than $1$, but I could not do this.

I am really confused as to what I am allowed to do to reason about projective curves. I would greatly appreciate any comments on my approach to this problem, particularly pertaining to whether it is correct to argue this way or what the major issues are with this argument. I know that there are probably some problems with my argument, so I apologise if some of the mistakes are quite grievous, I am just starting, so I am still uncertain about how to proceed.