This is true, but it's actually a bit subtle, because the completion $\hat A \to \hat B$ of an injective local homomorphism $A \to B$ of local domains need not be injective.
Given $f_1,\ldots,f_n \in k[x_1,\ldots,x_m]$ vanishing at the origin, we get a ring homomorphism $k[y_1,\ldots,y_n] \to k[x_1,\ldots,x_m]$ given by $y_i \mapsto f_i$. There are at least three different things one can mean by $k[[f_1,\ldots,f_n]]$ in this setting:
- a formal power series ring $k[[y_1,\ldots,y_n]]$ with a map $k[[y_1,\ldots,y_n]] \to k[[x_1,\ldots,x_m]]$ taking $y_i$ to $f_i$.
- the completion of the image $k[f_1,\ldots,f_n] \subseteq k[x_1,\ldots,x_m]$ at the origin.
- the image of either one of those in $k[[x_1,\ldots,x_m]]$. Because completion preserves surjections, it doesn't matter which of the two we take.
I will use the second as my definition of $k[[f_1,\ldots,f_n]]$, so we'll have to think a little bit about injectivity, but at least it's the completion of the subring $k[f_1,\ldots,f_n] \subseteq k[x_1,\ldots,x_m]$ that we're trying to understand. Despite these subtleties, we have:
Lemma. Let $k$ be a field, and let $A \hookrightarrow B$ be an injection of domains of finite type over $k$, where $B$ is normal. Let $\mathfrak n \subseteq B$ be a maximal ideal and $\mathfrak m = A \cap \mathfrak n$. Then the dimension of the image of $\hat A_{\mathfrak m} \to \hat B_{\mathfrak n}$ is the dimension of $A$.
Before giving a proof of the lemma, let's see how it implies the result you're after:
Corollary. Let $k$ be a field, and let $f_1,\ldots,f_n \in k[x_1,\ldots,x_m]$ be elements vanishing at the origin. If $g \in k[x_1,\ldots,x_m]$ is in the image of $k[[y_1,\ldots,y_n]] \to k[[x_1,\ldots,x_m]]$ given by $y_i \mapsto f_i$, then $g$ is algebraic over the subfield $k(f_1,\ldots,f_n)$ of $k(x_1,\ldots,x_m)$.
Proof. We may replace $g$ by $g-c$ for some $c \in k$ to assume $g$ vanishes at the origin as well. Write $A = k[f_1,\ldots,f_n]$, $B = k[f_1,\ldots,f_n,g]$, and $C = k[x_1,\ldots,x_m]$, and note that $C$ is normal (even smooth). The hypotheses imply that the images of $\hat A \to \hat C$ and $\hat B \to \hat C$ agree, so the lemma above shows that $\dim A = \dim B$. But $\dim A = \operatorname{trdeg}(k(f_1,\ldots,f_n)/k)$, and likewise for $B$ [Tag 00P0], so these coincide if and only if $k(f_1,\ldots,f_n) \to k(f_1,\ldots,f_n,g)$ is algebraic [Tag 030H]. $\square$
The following example shows the subtleties with completion and injections:
Example. Let $A = k[x,y]/(y^2 - x^3-x^2)$ be the affine coordinate ring of a nodal cubic (let's say $\operatorname{char} k \neq 2,3$), and let $B = k[t]$ be its normalisation via the map
\begin{align*}
k[x,y]/(y^2-x^3-x^2) &\to k[t] & &\\
x &\mapsto t^2+2t & &(= (t+1)^2-1),\\
y &\mapsto t^3+3t^2+2t & &(= (t+1)^3-(t+1)).
\end{align*}
Then the local homomorphism $A_{(x,y)} \to B_{(t)}$ is injective, but after completion it becomes isomorphic to
\begin{align*}
k[[x,y]]/(y^2-x^2) &\to k[[t]]\\
x, y &\mapsto 2t,
\end{align*}
which has kernel generated by $x-y$ (this is (partially) explained in Hartshorne, Chapter I, Example 5.6.3). Geometrically, this corresponds to the fact that $\operatorname{Spec} A$ has two branches $y = \pm x$ meeting at the origin, whereas $\operatorname{Spec} B$ only picks out the one corresponding to $y = x$ (indeed, $t+1 = y/x$, and we're zooming in around $t=0$).
The proof of the lemma above is 'standard' but tedious.
Proof of Lemma. This would be easy if $\hat A \to \hat B$ were injective, since $\dim \hat A = \dim A$ [Tag 07NV]. As in the case of the nodal curve above, we'll see that in general the map on completions $\hat A \to \hat B$ picks out certain local branches through $\mathfrak m$, which all have the same dimension.
By [Tag 0CB4(1)], there exists an étale $A$-algebra $A'$ and a prime $\mathfrak m' \subseteq A'$ above $\mathfrak m$ with $A/\mathfrak m \stackrel\sim\to A'/\mathfrak m'$ such that the number of local branches through $\mathfrak m$ is equal to the number of irreducible components of $A'$ through $\mathfrak m'$. Now $A'$ is no longer a domain, but it is still reduced [Tag 033B] and equidimensional. Let $B'$ be the tensor product $A' \otimes_A B$, with maximal ideal $\mathfrak n'$ corresponding to the surjection $B' \to A'/\mathfrak m' \otimes_{A/\mathfrak m} B/\mathfrak n$, which is just $B/\mathfrak n$ since $A/\mathfrak m \stackrel\sim\to A'/\mathfrak m'$ is an isomorphism. Then $B'$ is normal [Tag 033C], hence a product of normal domains [Tag 030C], so we replace $B'$ by the factor containing the prime $\mathfrak n'$ and assume $B'$ is a domain.
The maps $A \to A'$ and $B \to B'$ are étale and induce isomorphisms on residue fields, so they also induce isomorphisms on completions (this follows from [Tag 06LJ(b)], but should be easier). We have $\dim A = \dim A'$ and $\dim B = \dim B'$ [Tag 00ON], but $A'$ need not be a domain, nor $A' \to B'$ injective. The kernel $\mathfrak p$ of $A' \to B'$ is an ideal whose restriction to $A$ is $0$, and since $A \to A'$ is an injective map of equidimensional rings of the same dimension, this forces $\mathfrak p$ to be a minimal prime ideal. Let $A'' = A'/\mathfrak p \subseteq B'$, which by equidimensionality of $A'$ again has dimension $\dim A$.
Thus, we may replace $A \hookrightarrow B$ by $A'' \hookrightarrow B'$. Then all hypotheses of the lemma are still satisfied, and moreover $A$ has only one local branch at $\mathfrak m$. Then the same goes for $\hat A$ [Tag 0C2E], so $\hat A$ is unibranch [Tag 0C37(2)]. Since it is also reduced [Tag 07N2(3)], it is a domain [Tag 0BPZ]. By this answer, this implies that the map $\hat A \to \hat B$ is injective, which is the case we started with. $\square$
Picture. In the case of a nodal curve, the étale local picture separating the two branches through the origin is shown in Hartshorne, Chapter III, Exercise 10.6.
Best Answer
Set $f_i=x_i+x_4^2$ for $i=1,2,3$ and $f_4=x_1x_2+x_2x_3+x_3x_1$. Then $x_4^4=f_1f_2+f_2f_3+f_3f_1+f_4$, but no $x_i^2$ lies in the support of $f_4$.