Your proof seems to work, but it looks incomplete because you need to use more explicitly the fact that the ring of polynomials is an UFD.
The condition that the resultant $\mathrm{Res}(f,g)=0$ means that there are some polynomials $p,q$ such that $\deg p<\deg f$, $\deg q<\deg g$ and
$fq=gp$ (so $p/q=f/g$). In a UFD this is equivalent to having a common factor of $f$ and $g$.
To complete your approach you must replace the coefficient ring by its quotient field. You need the unique factorization at the end when you simplify the coefficients to be ring elements again.
*
For zero divisor free rings, without unique factorization, it will not work. For example, consider the polynomials $x^2-4y^2$ and $x^2+4xy+4y^2$. In some rings, like $(\mathbb{R}[y])[x]$, they have the common factor $x+2y$. In other rings, for example in $(\mathbb{Z}[4y,4y^2])[x]$, they don't.
However, the resultants will be the same zero polynomial in all rings.
So this method does not work if the coefficients are from the ring
$\mathbb{Z}[4y,4y^2]$.
*
Another possible approach is building a single (but huge) resultant. If $f,g\in\mathbb{R}[x_1,\dots,x_n]$ are nonzero polynomials then take all polynomials of the form $x_1^{a_1}\cdots x_n^{a_n}f$ with $a_1+\cdots+a_n<\deg g$ and $x_1^{b_1}\cdots x_n^{b_n}g$ with $b_1+\cdots+b_n<\deg f$. These polynomials are linearly dependent if and only if there are some polynomials $p,q$ with $\deg p<\deg f$, $\deg q<\deg g$ and $p/q=f/g$.
*
Sorry, I cannot recommend you books other than I can find by internet search...
:-(
This is just the Extended Euclidean Algorithm. Instead of back-substitution, I have always preferred to write the construction steps in the style of continued fractions. Furthermore, I have always depended on the kindness of strangers.
$$ \left( x^{3} - 6 x^{2} + 14 x - 15 \right) $$
$$ \left( x^{3} - 8 x^{2} + 21 x - 18 \right) $$
$$ \left( x^{3} - 6 x^{2} + 14 x - 15 \right) = \left( x^{3} - 8 x^{2} + 21 x - 18 \right) \cdot \color{magenta}{ \left( 1 \right) } + \left( 2 x^{2} - 7 x + 3 \right) $$
$$ \left( x^{3} - 8 x^{2} + 21 x - 18 \right) = \left( 2 x^{2} - 7 x + 3 \right) \cdot \color{magenta}{ \left( \frac{ 2 x - 9 }{ 4 } \right) } + \left( \frac{ 15 x - 45 }{ 4 } \right) $$
$$ \left( 2 x^{2} - 7 x + 3 \right) = \left( \frac{ 15 x - 45 }{ 4 } \right) \cdot \color{magenta}{ \left( \frac{ 8 x - 4 }{ 15 } \right) } + \left( 0 \right) $$
$$ \frac{ 0}{1} $$
$$ \frac{ 1}{0} $$
$$ \color{magenta}{ \left( 1 \right) } \Longrightarrow \Longrightarrow \frac{ \left( 1 \right) }{ \left( 1 \right) } $$
$$ \color{magenta}{ \left( \frac{ 2 x - 9 }{ 4 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 2 x - 5 }{ 4 } \right) }{ \left( \frac{ 2 x - 9 }{ 4 } \right) } $$
$$ \color{magenta}{ \left( \frac{ 8 x - 4 }{ 15 } \right) } \Longrightarrow \Longrightarrow \frac{ \left( \frac{ 4 x^{2} - 12 x + 20 }{ 15 } \right) }{ \left( \frac{ 4 x^{2} - 20 x + 24 }{ 15 } \right) } $$
$$ \left( x^{2} - 3 x + 5 \right) \left( \frac{ 2 x - 9 }{ 15 } \right) - \left( x^{2} - 5 x + 6 \right) \left( \frac{ 2 x - 5 }{ 15 } \right) = \left( -1 \right) $$
$$ \left( x^{3} - 6 x^{2} + 14 x - 15 \right) = \left( x^{2} - 3 x + 5 \right) \cdot \color{magenta}{ \left( x - 3 \right) } + \left( 0 \right) $$
$$ \left( x^{3} - 8 x^{2} + 21 x - 18 \right) = \left( x^{2} - 5 x + 6 \right) \cdot \color{magenta}{ \left( x - 3 \right) } + \left( 0 \right) $$
$$ \mbox{GCD} = \color{magenta}{ \left( x - 3 \right) } $$
$$ \left( x^{3} - 6 x^{2} + 14 x - 15 \right) \left( \frac{ 2 x - 9 }{ 15 } \right) - \left( x^{3} - 8 x^{2} + 21 x - 18 \right) \left( \frac{ 2 x - 5 }{ 15 } \right) = \left( - x + 3 \right) $$
............
Best Answer
Here is an explanation.
Let $f=x^3-x^2+3x-1$ and $g=x^3+2$.
Even though $\gcd(f,g)=1$ in $\mathbb{Z}[x]$, we cannot write $1=uf+vg$ with $u,v \in \mathbb{Z}[x]$ (because $\mathbb{Z}[x]$ is not a PID). But we can, if we allow $u,v \in \mathbb{Q}[x]$. Indeed, WA tells us that $$ 1 = \dfrac1{31} (-6 x^2-7 x-3)f(x)+\dfrac1{31}(6 x^2+x+14)g(x) $$ Therefore $$ 31 = (-6 x^2-7 x-3)f(x)+(6 x^2+x+14)g(x) $$ for all $x \in \mathbb Z$.
This only proves that $\gcd(f(x),g(x))=1$ or $31$ (because $31$ is prime), but it points to $31$.
So we try to solve $f(x)\equiv g(x) \equiv 0 \bmod 31$ and find that $x=27$ is a solution.
Therefore, $\gcd(f(x),g(x))=31$ for all $x=27+31k$. For all other values, $\gcd(f(x),g(x))=1$.