In advance, sorry for my english.
In this notes there's a multivariable version of the $|f(a)|_p<|f'(a)|_p$ version of Hensel's Lemma. I tried to adapt it to a proof for a multivariable version of the $f'(a)\ne 0\ \mod\ p$ Hensel's lemma. Basically, all I did was replace one univariate Hensel's lemma for the other, but as i'm new in the p adic world I am not so confident about it, I'd thank a lot for some comment.
Also, if someone knows of a book from somewhere where this is stated, I would thank the reference (if there is, I haven't found it). I'm mostly interested in this phenomena in ideal-adic version on some good ring.
So here is the statement, where $||\cdot||_p$ denotes the maximum of the $|\cdot|_p$ of a tuple, and $\nabla f$ is the tuple of partial derivatives of $f$
If $f(X_1 \dots X_d)\in\mathbb{Z}_p$ and some $\mathbf{a} \in\mathbb{Z}_p^d $ satisfies $$f(\mathbf{a})=0\ mod\ p$$ $$||\nabla f(\mathbf{a})||_p\ne 0$$ then $f$ has a zero in $\mathbb{Z}_p$ that equals $\mathbf{a}$ modulo $p$.
Here's the proof
As $||\nabla f(\mathbf{a})||_p\ne 0$, there's a coordinate $j$, fixed from now on, such that $|\partial f / \partial X_j(\mathbf{a})|_p\ne 0$. We now define a polynomial $g$ in one variable by $$g(X)=f(a_1,a_2,\dots,a_j-1,X,a_j+1,\dots,a_d)\in\mathbb{Z}_p[X].$$ Then $g(a_j)=f(\mathbf{a})$ and $g'(a_j)=\partial f / \partial X_j(\mathbf{a})$, so $g(a_j)=0\ mod\ p$ and $g'(a_j)\ne 0 \mod\ p$. Univariate Hensel's lemma can now be used, and there's $\alpha_j\in\mathbb{Z}_p$ such that $g(\alpha_j)=0$ and $\alpha = a_j \ mod\ p$. If we define $\alpha_i=a_i$ for $i\ne j$ and $\mathbf{\alpha}$ as $(\alpha_1,\dots ,\alpha_d)$, then $\mathbf{\alpha}\in\mathbb{Z}_p^d$ is a zero for $f$, and equals $\mathbf{a}$ modulo $p$.
:)