= Solution
Suppose there were distinct <singular points of an algebraic variety> $P,Q$. Let $L$ be their joining <projective line>. The homogeneous restriction $F|_L$ has degree $3$. At a <singular point of an algebraic variety> of a plane <algebraic curve>, the local defining equation has no constant or linear term, so its restriction to any line through that point vanishes to order at least $2$, unless the restriction is identically zero. Therefore a nonzero $F|_L$ would have at least four zeros counted with multiplicity, two at $P$ and two at $Q$, which a degree-$3$ homogeneous <polynomial> on $\mathbb P^1$ cannot have.
Hence $F|_L=0$. After choosing coordinates with $L=V(z)$, this says that every monomial of $F$ contains $z$, so $z$ divides $F$. That contradicts the <irreducible polynomial> hypothesis on the cubic. Thus
$$
\boxed{\text{An irreducible plane cubic has at most one singular point.}}
$$
This elementary version of the <Bézout theorem> argument works in every <characteristic>, without assuming that the <partial derivatives> of the cubic have their characteristic-zero form.
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