First a comment about projective spaces $\mathbb{P}(V)$, where $V$ is any vector space over a field $k$: the usual definition of this object is that of lines in $V$ passing through the origin. Every such a line $\ell$ can be identified just by giving a nonzero vector $v\in \ell$: indeed this vector itself spans a $1$-dimensional subspace of $\mathbb{C}^n$ that is contained in the $1$-dimensional subspace $\ell$, so that it must coincide with $\ell$ itself.
Moreover, two nonzero vectors $v_1,v_2 \in V\setminus \{0\}$ determine the same line if and only they span the same subspace, i.e. if and only if they are linearly dependent, that is there is a nonzero scalar $\lambda \in k, \lambda\ne 0$ such that
$$ v_1 = \lambda \cdot v_2 $$
Hence, the set $\mathbb{P}(V)$ can be identified with $V\setminus \{ 0 \}$ quotiented by the equivalence relation
$$ v_1 \sim v_2 \iff v_1 = \lambda\cdot v_2 \text{ for a certain } \lambda \in k, \lambda \ne 0 $$
(it is easy to show that the above is actually an equivalence relation).
Now, by definition $\mathbb{P}^n(\mathbb{C}):=\mathbb{P}(\mathbb{C}^{n+1})$, so that it can be considered as the set of nonzero $n+1$-tuples $(a_0,a_1,\dots,a_n)\in \mathbb{C}^{n+1}\setminus \{0 \}$ (because these are exactly the nonzero vectors in $\mathbb{C}^{n+1}$) quotiented by the above equivalence relation. In particular, if we denote the class of equivalence of the nonzero $n+1$-tuple $(a_0,\dots,a_n)$ by $[a_0,\dots,a_n]$ we see that
$$ \mathbb{P}^n(\mathbb{C})=\{ [a_0,\dots,a_n]\,\, \mid \,\, (a_0,\dots,a_n)\in \mathbb{C}^{n+1}\setminus \{0\} \} $$
so that there is no point with all zero coordinates in $\mathbb{P}^n(\mathbb{C})$.
Moreover, we have the equality in $\mathbb{P}^n(\mathbb{C})$
$$ [a_0,\dots,a_n]= [\lambda\cdot a_0,\dots,\lambda\cdot a_n] $$
for every $(a_0,\dots,a_n) \in \mathbb{C}^{n+1}\setminus \{0\}$ and $\lambda\in \mathbb{C},\lambda\ne 0$. Indeed, we have that
$$ (a_0,\dots,a_n) \sim (\lambda\cdot a_0,\dots,\lambda\cdot a_n)$$
in $V\setminus \{ 0\}$.
To compute the flexes, first we prove that the curve has no singular points: a singular point is a point on the curve where all the partail derivatives $\frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z}$ vanish. Computing the partial derivatives we see that
$$ \frac{\partial f}{\partial x} = 3x^2 \qquad \frac{\partial f}{\partial y} = 3y^2 \qquad \frac{\partial f}{\partial z} = 3z^2 $$
and then they cannot vanish all together since, by definition, the point [0,0,0] does not belong to the projective space $\mathbb{P}^2(\mathbb{C})$.
Now, since every point of the curve is nonsingular we just have to compute the intersection of the curve with the one given by the Hessian. As you say the Hessian is equal to $216 \cdot xyz$, and for this to be zero one between $x,y,z$ must be zero.
For example suppose that $z=0$, then the condition for a point $[x,y,0]$ to be on the curve is that
$$ x^3 + y^3 = 0 $$
Now we observe that it cannot be that $x=0$ or $y=0$: indeed, if for example $x=0$ then the equation above tells us that $y=0$ as well, but as before the point $[0,0,0]$ is not a point in $\mathbb{P}^2(\mathbb{C})$. The same reasoning holds if $y=0$.
Now, since it must be that $x\ne 0$ and $y\ne 0$ we can suppose that $y=-1$: indeed, by definition the point $[x,y,0]$ is equal to the point $[\frac{x}{-y},-1,0]$ and then the equation that we have to solve becomes
$$ x^3-1=0 $$
the solutions of this equation are given by
$$ x=1,\,\, x=e^{\frac{2\pi}{3}i}, \,\, x=e^{\frac{4\pi}{3}i} $$
so that this gives us three flexes of the curve
$$ [1,-1,0], \qquad [e^{\frac{2\pi}{3}i},-1,0] \qquad [e^{\frac{4\pi}{3}i},-1,0] $$
The situation in which $y=0$ or $x=0$ are symmetric to this one and then we find that the other flexes are
$$ [1,0,-1], \qquad [e^{\frac{2\pi}{3}i},0,-1] \qquad [e^{\frac{4\pi}{3}i},0,-1] $$
$$ [0,1,-1], \qquad [0,e^{\frac{2\pi}{3}i},-1] \qquad [0,e^{\frac{4\pi}{3}i},-1] $$
and these nine points are all the flexes of the curve.
