Ring Isomorphism Between Field of Fractions 
If $R\DeclareMathOperator{\Frac}{Frac}$ is a domain with $F = \Frac(R)$, prove that $\Frac(R[x]) \simeq F(x)$.

The way I solved this problem was by defining $f : \Frac(R[x]) \to F(x)$ by $f([h(x),g(x)]) = \frac{h(x)}{g(x)}$, which is sensible since $g(x) \neq 0$. I then proceeded to show that it is well-defined, a ring homomorphism, and bijective.
But I am wondering whether there is a way to use the following theorems to cut down on some of the work:

(1) If $R$ is a field, then $R \simeq \Frac(R)$. More precisely, show that the ring homomorphism $f : R \to \Frac(R)$, given by $r \mapsto [r,1]$, is an isomorphism
(2) If $A$ and $R$ are domains and $\varphi : A \to R$ is a ring isomorphism, then $[a,b] \mapsto [\varphi(a), \varphi (b)]$ is an isomorphism.
(3) Let $R$ be a domain and let $\varphi : R \to k$ be an injective ring homomorphism, where $k$ is a field. Prove that there exists a unique ring homomorphism $\phi : \Frac(R) \to k$ extending $\varphi\,$; i.e., $\phi |_{R} = \varphi$.

I found a solution online which seems to use them, although it was so poorly written that I had, for the most part, trouble following it. I have tried to think about how to use these theorems, but it seems that using them will, in the end, amount to the same thing I did already. Can you see anyway of using them?

Hint: (a). Show that there is an injective ring homomorphism $R[x] \longrightarrow F(x)$. Then use (3) to get a map from $Frac(R[x]) \longrightarrow F(x)$.
(b) Show that there is an injective ring homomorphism $F[x] \longrightarrow Frac(R[x])$. Again use (3) to get a map from $F(x) \longrightarrow Frac(R[x])$.
(c) Use the uniqueness property stated in (3) to show that the above maps are inverse to each other.
