I'm familiar with Hensel's Lemma in the case where the polynomial under consideration is monic (or has invertible leading coefficient), but I'm trying to understand how it works in the case where the leading coefficient is $0$ modulo the prime one is working with. Here is an example I constructed:
Let $F(x,z) = 3x^3 + ax^2z + bxz^2 + cz^3$. Suppose that $F(x,z)$ has no repeated roots in $\mathbb{P}^1(\mathbb{F}_3)$, and further suppose that $F(x,1) \equiv 0 \pmod 9$ for each $x \in \{1,4,7\}$. I claim that this yields a contradiction to Hensel's Lemma. Indeed, $[x : z] = [1 : 1]$ is a non-repeated root of $F(x,z)$ modulo $3$ that has $3$ distinct lifts to $\mathbb{Z}/9\mathbb{Z}$, namely $(x,z) = (1,1)$, $(4,1)$, and $(7,1)$. But Hensel's Lemma says that because $[x :z] = [1 : 1]$ is a non-repeated root, there can only be one lift.
Have I made an error in the above argument? Also, I think the criterion for $F(x,z)$ not to have repeated roots is that the partials $F_x(x,z)$ and $F_z(x,z)$ should not have any common roots.