Isomorphism between $\Bbb R$ and $\Bbb R(X)$?

There is no homomorphism $\mathbb{R}(X)\to\mathbb{R}$. Indeed, the only homomorphism $\mathbb{R}\to\mathbb{R}$ is the identity, so a homomorphism $\mathbb{R}(X)\to\mathbb{R}$ would have to restrict to the identity on constants and then there is nowhere $X$ could go.

Here's a proof that the only homomorphism $f:\mathbb{R}\to\mathbb{R}$ is the identity. We know $f$ must be the identity on $\mathbb{Q}$. Now note that for any $r\in\mathbb{R}$ and any rational $q<r$, we must have $$f(\sqrt{r-q})^2=f(r-q)=f(r)-q,$$ so $f(r)-q\geq 0$ since it is a square. This means $f(r)\geq r$ for all $r$. By a similar argument using $\sqrt{q-r}$ instead of $\sqrt{r-q}$, you can also show that $f(r)\leq r$. Thus $f(r)=r$ for all $r$.