= Solution
Over $\overline{\mathbb F}_3$, take the projective cubic
$$
\boxed{C=V_+(Y^2Z-X^3+XZ^2)\subseteq\mathbb P^2.}
$$
In characteristic three its partial <derivatives> are $F_X=Z^2$, $F_Y=2YZ$ and $F_Z=Y^2+2XZ$. If they vanished simultaneously, $Z=0$ and then $Y=0$. The equation of $C$ would force $X=0$, impossible for a projective point. Thus the <Jacobian criterion> proves smoothness everywhere, including infinity.
It is an irreducible cubic: on the affine chart $Z=1$, the polynomial is $y^2-(x^3-x)$, and $x^3-x$ is not a square in $\overline{\mathbb F}_3(x)$ because its pole at infinity has odd order three. A quadratic in $y$ would factor over that field only if its right side were a square. <Gauss lemma for polynomials> then gives affine irreducibility, and homogenization introduces no factor at infinity because $Z$ does not divide $F$. This supplies the <smooth plane cubic in characteristic three>. In particular, choosing a Fermat cubic here would fail, because its cubic partial <derivatives> vanish in characteristic three.
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