The picture of $\{(x:y:z) \in \mathbb P_{\mathbb C}^2 | yz =0\}$ is two spheres (each representing a copy of $\mathbb P_{\mathbb C}^1$) intersecting at one point (representing $(1,0,0)$).
But
why is the picture of $\{(x:y:z) \in \mathbb P_{\mathbb C}^2 | x^3 + y^3 = xyz \}$ a two-dimensional sphere with two points identified? Moreover, is there any computer program that can determine the picture of a projective variety?
Your equation $ x^3 + y^3 = xyz$ defines an algebraic complex subvariety $C\subset \mathbb P^2(\mathbb C)$ of complex dimension $1$, hence called a complex curve, but of real dimension $2$.
This complex curve has a single nodal singularity at $P=[0:0:1]$, the only point where the curve is not a real manifold of dimension $2$.
Algebraic geometers have a canonical procedure for eliminating such a singularity, called normalization.
In the concrete case at hand the normalization is the algebraic morphism $$ \nu:S^2=\mathbb P^1(\mathbb C)\to C:[t:u]\mapsto [tu^2:t^2u:u^3+t^3] $$ This morphism is surjective and almost injective.
More precisely the points $P_1=[0:1]$ and $P_2=[1:0]$ both get sent to $P$ and the other points of $S^2$ are mapped isomorphically onto $C\setminus \{P\}$.
The normalization $\nu$ is thus the identification map you requested.
