= Solution
Writing $X=u$, $Y=v$, and $Z=-w$ turns the cubic into $X^3+Y^3=Z^3$. The birational coordinates
$$
x=\frac{12Z}{X+Y},
\qquad y=\frac{36(X-Y)}{X+Y}
$$
give the Weierstrass equation
$$
y^2=x^3-432.
$$
The original projective cubic has no common zero of its three partial derivatives modulo any $p\ne3$, so it is already a smooth proper model and $C$ has <reduction of an elliptic curve>[good reduction] at every such prime.
If $p\equiv1\pmod3$, then $\mu_3\subset\mathbb F_p$, all nine inflection points are rational, and $C(\mathbb F_p)$ contains $(\mathbb Z/3\mathbb Z)^2$, so it is not cyclic. If $p\equiv2\pmod3$, cubing is a bijection on $\mathbb F_p$. Counting on the Fermat model gives $\#C(\mathbb F_p)=p+1$. A finite elliptic-curve group has the form $\mathbb Z/d\mathbb Z\times\mathbb Z/e\mathbb Z$ with $d\mid e$ and $d\mid p-1$, so $d\mid\gcd(p-1,p+1)=2$. For odd $p$, the equation $x^3=432$ has exactly one root because cubing is bijective, so there is only one nonzero rational 2-torsion point and $d\ne2$. Hence the group is cyclic. For $p=2$ it has order three and is cyclic as well.
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