The user94293's answer is correct, If you look at the equation you will see that the result will be zero if you add the number itself. It is designed for the points $P$ and $Q$ such that $P\neq Q$. If you want to calculate the $P+P$ that is called point doubling and it is written as $[2]P$. There are always special formulas for doubling since it adds the point itself.
The addition formula is optimized and cannot be used for $2[p]$. From your questions and comments, I understand that you are trying to find the scalar multiplication by just adding;
Scalar Multiplication
The scalar multiplication $[k]P$ this actually means adding $P$, $k$-time itself. More formally;
let $k \in \mathbb{N}\backslash\{ 0\}$
\begin{align}
[k]:& E \to E\\
&P\mapsto [k]P=\underbrace{P+P+\cdots+P}_{\text{$k$ times}}.\end{align}
Double-and-Add
To use these formulas correctly, you need the double-and-add algorithm
#Returns [k]G, where p is modulus
#and c is the constant of the Generalized Hessian Curve
doubleAndAdd(G, k , p ,c)
N = P
Q = 0
for i in bits(k)
if i = 1 then
Q = point_add(Q, N)
N = point_double(N)
return Q
Where the bits convert the $k$ into binary form and started from LSB.
Updates on implementation
I've found a problem during the calculation of the order an element, later figured out the cause. One can reach the codes and explanations on the Q/A from the StackOverflow
The order of $(4,2,6)$ is $77400$ i.e. $[77400](4,2,6) = \mathcal{O}$ and the order factors into $2,2,2,3,3,5,5,43$. Not a good curve.
To run better, the order of the curve must be calculated before and so that one can use $[k]P = [k \bmod order]P$ equality.
Smart's Sample Curve
N.Smart wrote an article titled The Hessian Form of an Elliptic Curve. In this work first, a curve over $\operatorname{GF}(2^{191})$ on the form $$E': y^2 +xy = x^3 + x^2 + b$$ $$b = \texttt{0x4DE3965E00F2A1C6C9750156A6FEFBE5EEF780BF3EF20E48}$$ is given, and that then transferred to a Hessian form.
$$x^3 + y^3 + z^3 = Dxyz$$
$$D = \texttt{0x16A4C7C2030FAD1380ABF8C2D47DC3E0C20AF62F6EDD06A7}$$
A point of order $q$ on $C$ is also given by $(x, y, z)$, where
$$x = \texttt{0x52FD0CE78D0651B4F66D2F4E12E170CA3E429F6A06433B22}$$
$$y = \texttt{0x1BECA50368403F3D13173968082B035397C77830A9D90E5D}$$
$$z = \texttt{0x2B08F7C0CCAC86151AA6FECABDD2D052BD60924F28A6A78E}$$
That is not generalized, however, your $c=1$ make is in this case. You can benefit from this curve.
