Let $\mathbb{F}=\mathbb{Q}(x,y)$ and $\mathbb{K}=\mathbb{Q}(x)$ be the fields of rational functions in $x,y$ and $x$, respectively.
Let $A$ be the subring of $\mathbb{F}$ generated by $x,y,\frac{x+1}{y},\frac{y+1}{x},\frac{x+y+1}{xy}$ and $B$ be the subring of $\mathbb{K}$ generated by $x,\frac{2}{x}$.
The map $f:A \rightarrow B$ sends $x$ to itself, $y$ to $\frac{x}{2}$ and the remaining generators from $A$ to $1$ in $B$.
I want to show that this map is a ring homomorphism.
Now, $f$ sends generators of $A$ to elements in $B$ so we have that $B \subset f(A)$. Next we want to show that $f$ preserves the addition and multiplication. Let $u,v \in A$. Then $u$ and $v$ are polynomials in the generators of $A$ and so we write $u=p(\textbf{x})$ and $v=q(\textbf{x})$, where $\textbf{x}$ denotes the set of the generators of $A$. Then, \begin{equation} f(u+v)=f(p(\textbf{x})+q(\textbf{x}))=f((p+q)(\textbf{x}))=(p+q)(f(\textbf{x}))=p(f(\textbf{x})) + q(f(\textbf{x})) = f(p(\textbf{x}))+f(q(\textbf{x})) = f(u)+f(v). \end{equation} Similarly, \begin{equation} f(uv)=f(p(\textbf{x})q(\textbf{x}))=f((pq)(\textbf{x}))=(pq)(f(\textbf{x}))=p(f(\textbf{x}))q(f(\textbf{x})) = f(p(\textbf{x}))f(q(\textbf{x})) = f(u)f(v). \end{equation} Finally, $f(1)=1$ and so $f$ is a ring homomorphism, as required. Is this correct?
Here is an easy argument. The ring $A$ is actually a subring of the ring $\mathbb Q[x^{\pm1},y^{\pm1}]$ of Laurent polynomials. To define a map $\mathbb Q[x^{\pm1},y^{\pm1}]\to\mathbb K$, one just needs to specify the images of $x$ and $y$, which must be invertible. So your map $f$ determines a ring homomorphism $\mathbb Q[x^{\pm1},y^{\pm1}]\to\mathbb K$, and hence restricts to a ring homomorphism $A\to\mathbb K$.