I think it's well known that if $X\subset\mathbb{P}^3$ is a smooth cubic surface and we take the projection $\pi: X\rightarrow \mathbb{P}^2$ from a point off the surface, then it's branched over a sextic curve with 6 cusps. 


Why is this true? In particular, I'm not seeing the 6 cusps. 




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Example of a statement of the fact: *the Problem of Existence of Algebraic Functions* by Zariski (1929), pg 320



Branch locus of projection of cubic surface
A slightly simpler way: if you project from $(0,0,0,1)$, after a change of coordinates you can write the equation of your surface as $T^3+PT+Q=0$, where $P$ and $Q$ are forms of degree $2$ and $3$ in $X,Y,Z$. The branch curve is given by $4P^3+27Q^2=0$, and it is fairly easy to see that the 6 points given by $P=Q=0$ are cusps. 


