I believe that for any subfield $F$ of $\Bbb R$, with finite degree of transcendence over $\Bbb Q$, the field of rational fractions in one variable $F(x)$ can't be embedded in $F$. However the situation can be different for those with infinite degree of transcendence over $\Bbb Q$.
I know that $F(x)$ always embeds in $\Bbb C$. Actually $\Bbb C(x)$ embeds in $\Bbb C^{[1]}$. I only know that if $j:F(x) \hookrightarrow F$ denotes a field morphism, then $j(x)$ is transcendental over $\Bbb Q$.
Here is an idea for $F=\Bbb R$. Let's take a transcendence basis $T$ of $\Bbb R$ over $\Bbb Q$ (which has cardinality $2^{\aleph_0}$) and complete it into a (linear algebra) basis $B$ of $\Bbb R$ over $\Bbb Q$. Then $T$ and $T \cup \{x\}$ ($x$ is the indeterminate) have the same cardinality, we can pick a bijection $b:T \cup \{x\}\to T$. Then we could try to build some $j : \Bbb R(x)=\Bbb Q(B \cup \{x\}) \to \Bbb Q(B)=\Bbb R$...
Any hint would be appreciated!
$^{[1]}$ A nice example from heptagon shows the existence of a subfield $L \subset \Bbb C$ such that $L(x) \cong L$. Indeed, we can embed $L=\Bbb R(x_1,x_2,\dots)$ in $\Bbb C$.
I think I've found an example, thanks to this answer. We pick a subset $\{x_i\}_{i\in \mathbb{N}}$ of pairwise distinct elements of a transcendence basis of $\mathbb{R}$ over $\mathbb{Q}$ (this is possible since a transcendence basis of $\Bbb R$ over $\Bbb Q$ has cardinality $2^{\aleph_0}$).
Then the subfield $F = \Bbb Q(x_1, \dots, x_n, \dots)$ satisfies $F \cong F(X) \cong F(X,Y) \cong \cdots$ (as fields), so in particular $F(X)$ embeds in $F$.