Evaluate $\frac{1}{zx+y-1}+\frac{1}{zy+x-1}+\frac{1}{xy+z-1}$ if $x+y+z=2,x^2+y^2+z^2=3,xyz=4$
If we transform the whole thing into a single fraction, then both numerator and denominator will be symmetric polynomials on three variables.
The good thing about them is that one can then use the Fundamental Theorem of Symmetric Polynomials to write $P(x,y,z)$ in a unique way as a polynomial $P'(e_1(x,y,z),e_2(x,y,z),e_3(x,y,z))$ where $$e_1(x,y,z) = x + y + z \qquad e_2(x,y,z) = xy + xz + yz \qquad e_3(x,y,z) = xyz.$$
We already know that $e_1(x,y,z) = 2$ and $e_3(x,y,z) = 4$. We can compute $e_2$ since $e_1^2 - (x^2 + y^2 + z^2) = 2e_2$: $$e_2 = \dfrac{1}{2}.$$
Now the only thing we have to do is to decompose the numerator and denominator as a polynomial on $e_1,e_2,e_3$ to find the value of the initial problem. This, however, may not be very elegant.
The whole thing gives $P/Q$ where:
$$P = (x^2yz+xy^2z+xyz^2) + (x^2y+x^2z+xy^2+y^2z+xz^2+yz^2) - (xy+xz+yz) - (2x+2y+2z) + 3,$$
$$Q = (x^3yz+xy^3z+xyz^3) + (x^2y^2z^2) + (x^2y^2+x^2z^2+y^2z^2) -( x^2yz+xy^2z+xyz^2) - (x^2y+x^2z+xy^2+y^2z+xz^2+yz^2) + (x+y+z) + (xyz)- 1.$$
Therefore
$$x^2yz+xy^2z+xyz^2 = xyz(x+y+z) = 8$$ $$x^2y+x^2z+xy^2+y^2z+xz^2+yz^2 = -3xyz + (xy+xz+yz)(x+y+z) = -11$$ $$2x+2y+2z = 2(x+y+z) = 4$$
So $P = 8 - 11 - \dfrac{1}{2} - 4 + 3 = -\dfrac{9}{2}$.
Conversely,
$$x^3yz + xy^3z + xyz^3 = (xyz)(x^2 + y^2 + z^2) = 12$$ $$x^2y^2z^2 = (xyz)^2 = 16$$ $$x^2y^2+x^2z^2+y^2z^2 - x^2yz-xy^2z-xyz^2 = (xy+xz+yz)^2 - 3(xyz)(x+y+z) = -\dfrac{95}{4}$$ $$x^2y+x^2z+xy^2+y^2z+xz^2+yz^2 = -3xyz + (xy+xz+yz)(x+y+z) = -11$$
so $Q = 12 + 16 -\dfrac{95}{4} + 11 + 2 + 4 - 1 = \dfrac{81}{4}$.
Finally, $P/Q = -\dfrac{9\cdot 4}{2\cdot 81} = -\dfrac{2}{9}.$
Edit : Since the OP changes some signs, this answer also changes the signs.
You have already noticed that $$\frac{1}{zx+y-1}+\frac{1}{yz+x-1}+\frac{1}{xy+z-1}=\frac{y}{y^2-y+4}+\frac{x}{x^2-x+4}+\frac{z}{z^2-z+4}$$ This answer shows that using that $x,y,z$ are the solutions of $t^3-2t^2+\frac 12t-4=0$ enables us to have a simpler form and to find the value easily.
We have that $$x+y+z=2,\quad xyz=4$$ and that $$xy+yz+zx=\frac 12\left((x+y+z)^2-(x^2+y^2+z^2)\right)=\frac{2^2-3}{2}=\frac 12$$
So, we know that $x,y,z$ are the solutions of $$t^3-2t^2+\frac 12t-4=0,$$ i.e. $$(t-1)(t^2-t+4)-\frac{9}{2}t=0,$$ i.e. $$\frac{t}{t^2-t+4}=\frac{2}{9}(t-1)$$ since $t^2-t+4\not=0$.
Hence, we get $$\begin{align}\frac{1}{zx+y-1}+\frac{1}{yz+x-1}+\frac{1}{xy+z-1}&=\frac{y}{y^2-y+4}+\frac{x}{x^2-x+4}+\frac{z}{z^2-z+4}\\\\&=\frac{2}{9}(y-1)+\frac{2}{9}(x-1)+\frac{2}{9}(z-1)\\\\&=\frac{2}{9}(x+y+z-3)\\\\&=\color{red}{-\frac{2}{9}}\end{align}$$
Replacing $x=2-y-z$ and similar, the sum becomes:
$$ S = \sum_{cyc}\frac{1}{x+yz-1} = \sum_{cyc}\frac{1}{1-y-z+yz} = \sum_{cyc}\frac{1}{(y-1)(z-1)}=\cfrac{1}{\prod_{cyc} (x-1)} \sum_{cyc} (x-1) $$
From the given relations $\sum_{cyc} xy = \frac{1}{2}\left((\sum_{cyc} x)^2-\sum_{cyc} x^2\right)=\cfrac{1}{2}\,$, so:
$$ \begin{align} \sum_{cyc}(x-1) & = \sum_{cyc} x - 3 = 2 - 3 = -1 \\ \prod_{cyc}(x-1) & = xyz - \sum_{cyc} xy + \sum_{cyc} x - 1 = 4 - \cfrac{1}{2} + 2 - 1 = \cfrac{9}{2} \end{align} $$
Therefore $\,S=\cfrac{1}{\frac{9}{2}} \cdot (-1) = - \cfrac{2}{9}\,$.
P.S. For an alternative derivation, note that $x,y,z$ are the roots of $\,2t^3-4t^2+t-8=0$ by Vieta's formulas. The polynomial with roots $x-1,y-1,z-1$ can be obtained with the substitution $t=u+1$ which, after expanding the powers and collecting, results in $2u^3+2u^2-u-9=0\,$. Therefore $\,\sum_{cyc} (x-1) = -1\,$ and $\,\prod_{cyc} (x-1) = \cfrac{9}{2}\,$.