Will this work as a proof?
Let $R$ be a domain and $L$ a field. Let $f : R \to L$ be a ring hom. Let $K$ be the field of fractions of $R$. Then to extend $f$ to $K$ means there is a map $f^* : K \to L$ such that $f^*|_R = f$. The obvious choice for $f^*$ is that it respects precisely the fractions so that $f^*(a/b) = f(a)f(b)^{-1}$. Let $g : K \to L$ be a ring hom that also restricts to $f$ on $R$. Then $g(a/b) = g(a \cdot (1/b)) = g(a) \cdot g(1/b)$, but $1/b \equiv b^{-1}$ in field $K$ so that $g(a/b) = g(a)g(b)^{-1} = f(a) f(b)^{-1} = f^*(a/b). \ \ \ \ \square$
Thank you. I am putting this into a flash card so don't want to write down an invalid proof.
You need to check the map is well defined. What does this mean? It means if we select two equivalent elements of the field of fractions $a/b \sim c/d$ then your definition of $f(a/b)$ and $f(c/d)$ give the same answer.