Etale iff Completions are Isomorphic

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I have a question aboutthe proof of proposition 4.3.26 from Liu's "Algebraic Geometry and Arithmetic Curves":

Let $Y$ be a locally Noetherian scheme and $f:X \to Y$ a morphism of finite type. Fix a point $x$ and denote $y=f(x)$. Assume that $O_{X,x}/m_x=k(x)=k(y)=O_{Y,y}/m_y$ coincide.

The local ring morphism on stalks $f_x:O_{Y,y} \to O_{X,x}$ induces a morphism of completions

$$\widehat{f}_x:\widehat{O}_{Y,y} \to \widehat{O}_{X,x}$$

where the completions are induced wrt to corresponding maximal ideals $m_x$ and $m_y$.

The proposition says that $f$ is etale in $x$ if and only if $\widehat{f}_x:\widehat{O}_{Y,y} \to \widehat{O}_{X,x}$ is an isomorphism.

The "$\Rightarrow$" I understand. Regarding the "$\Leftarrow$" I don't understand the argument.

Liu refers to Theorem 1.3.16 and Exercise 1.2.18.

Recall etale means flat and unramified.

Exercise 1.2.18 solves the flatness of $f_x$: It says that if $B \to E$ is faithful flat then $A \to B$ is flat if and only if $A \to E$ is flat. Taking for $A \to B$ the morphism $f_x$ and for $E$ the completion we are done since completions of local rings are faithful flat.

Recall unramified means that $k(y) \to k(x)$ is separable and $m_yO_{X,x}=m_x$.

QUESTION: By assumption $k(y) = k(x)$ so the first one is ok but what about $m_yO_{X,x}=m_x$?

My considerations:Since $\widehat{f}_x$ is an iso we deduce that

  1. $\widehat{m_x} = \widehat{m_y}\widehat{O}_{X,x}$ as completions

  2. $O_{Y,y}/m_y \cong \widehat{O}_{Y,y}/\widehat{m_y} \to \widehat{O}_{X,x}/\widehat{m_x} \cong O_{X,x}/m_x$ are surjective.

  3. for every $k$ there exist a $d_k$ with $ g(m_y)^{d_k} \subset m_x ^k$ and vice versa (so same topology)

But I don't see how to conclude that $m_yO_{X,x}=m_x$.