Tangent lines of a projective curve I am solving this exercise and I don't know if I did the last step in a correct way:

Prove that the projective curve $(X+Y+Z)^{3}=27XYZ$ is irreducible and compute its singular points with their multiplicities and tangent lines.

Let $F=(X+Y+Z)^{3}-27XYZ$. As $Z$ does not divide $F$, dehomogenizing is reversible. The dehomogenized polynomial is: $$F_{*}=(X+Y+1)^{3}-27XY$$ $F_{*}$ is the substraction of two forms of consecutive degrees without common divisors, so it it irreducible. If $F_{*}$ is irreducible, then $(F_{*})^{*}=F$ is irreducible too.
I have computed the singularities using Maple and I get that $P=(1:1:1)$ is a double point.
Now there is the question. Have I computed well the tangent lines? I have done the following:
First I find the derivatives:$$F_{X}=3(X+Y+Z)^{2}-27YZ,\quad F_{Y}=3(X+Y+Z)^{2}-27XZ,\quad F_{Z}=3(X+Y+Z)^{2}-27XY$$ $$F_{XX}=6(X+Y+Z),\quad F_{XY}=6(X+Y+Z)-27Z,\quad F_{XZ}=6(X+Y+Z)-27Y$$ $$F_{YY}=6(X+Y+Z),\quad F_{YZ}=6(X+Y+Z)-27X,\quad F_{ZZ}=6(X+Y+Z)$$
And I will use the equation of the tangent lines: $$\sum_{i+j+k=m}\frac{1}{i!j!k!}\frac{\partial^{m}F}{\partial X^{i}\partial Y^{j}\partial Z^{k}}(P)X^{i}Y^{j}Z^{k}=0$$
Then: $$\frac{1}{2!}(F_{XX}(P)X^{2}+F_{YY}(P)Y^{2}+F_{ZZ}(P)Z^{2})+\frac{1}{1!}(2F_{XY}(P)XY+2F_{YZ}(P)YZ+2F_{XZ}(P)XZ)=0$$
So I get that: $$3X^{2}+3Y^{2}+3Z^{2}-18XY-18YZ-18XZ=0$$
Is this the equation of the tangent lines?
 A: $F_{XX}(P)=18\ne 3$ etc. The final conic should be
$$18X^2+18Y^2+18Z^2-18XY-18XZ-18YZ=0,$$
equivalently,
$$X^2+Y^2+Z^2-XY-XZ-YZ=0,$$
which is
$$(X+\omega Y+\omega^2 Z)(X+\omega^2 Y+\omega Z)=0$$
with $\omega=\exp(2\pi i/3)$.
A: Notice that
$$\{ (i, j, k) \mid i + j + k = 2 \} = \{(2, 0, 0), (0, 2, 0), (0, 0, 2), (1, 1, 0), (0, 1, 1), (1, 0, 1) \}$$
Thus we have
$$\frac{1}{2}(F_{XX}(P)X^{2}+F_{YY}(P)Y^{2}+F_{ZZ}(P)Z^{2})+(F_{XY}(P)XY+F_{YZ}(P)YZ+F_{XZ}(P)XZ) = 0$$
without the coefficient $2$ multiplying the mixed partial derivatives. It might be easier to understand why the coefficient $\frac 1 2$ must be there instead if you write the equation as:
$$(F_{XX}(P)X^{2}+F_{YY}(P)Y^{2}+F_{ZZ}(P)Z^{2})+2(F_{XY}(P)XY+F_{YZ}(P)YZ+F_{XZ}(P)XZ) = 0$$
because in this way one sees that the coefficient $2$ is there exactly to account for both $F_{XY}(P)XY$ and $F_{YX}(P)YX$, etc.
Anyway, a different approach is the following. Since $P$ belongs to the affine open subset given by $Z \neq 0$, we can dehomogenize with $Z = 1$ and identify $P$ with $(1, 1)$. Then, we consider the affinity:
$$\begin{cases} x = X - 1 \\ y = Y - 1 \end{cases} \quad\Longleftrightarrow\quad \begin{cases}X = x + 1 \\ Y = y + 1 \end{cases}$$
which maps $P = (1, 1)$ to $O = (0, 0)$. Applying the affinity to $F_*$ gives us:
$$(x + y + 3)^3 - 27(x + 1)(y + 1) = x^3 + 3x^2y + 3xy^2 + y^3 + 9x^2 - 9xy + 9y^2.$$
The homogeneous component with the lowest degree is:
$$9(x^2 - xy + y^2) = 9 (x - \omega y)(x - \omega^2 y)$$
where $\omega$ is a primitive sixth root of $1$. Thus the tangent lines to the transformed curve in $O$ have equations:
$$x - \omega y = 0 \qquad x - \omega^2 y = 0.$$
We apply the inverse affinity to obtain the equations of the tangent lines to the initial curve in $P$:
$$(X - 1) - \omega (Y - 1) = 0 \quad\longrightarrow\quad X - \omega Y - \omega^2 = 0$$
$$(X - 1) - \omega^2 (Y - 1) = 0 \quad\longrightarrow\quad X - \omega^2 Y - \omega = 0$$
Finally, we homogenize the equations:
$$X - \omega Y - \omega^2 Z = 0 \qquad X - \omega^2 Y - \omega Z = 0.$$
