AM-GM may help

$\displaystyle\begin{cases}x+\frac{1}{yz}+z^2=3\\y+\frac{1}{xz}=2\end{cases}\implies x+\frac{1}{yz}+z^2+y+\frac{1}{xz}=5\stackrel{\text{AM-GM}}\ge 5\sqrt[5]{x\cdot \frac{1}{yz}\cdot z^2\cdot y\cdot \frac{1}{xz}}=5$

The minimum value of $x+\frac{1}{yz}+z^2+y+\frac{1}{xz}$ is $5$ and it is reached if and only if $x=\frac{1}{yz}=z^2=y=\frac{1}{xz}$ .

Since we know $x+\frac{1}{yz}+z^2+y+\frac{1}{xz}$ is equal to its minimum value, we have $x=\frac{1}{yz}=z^2=y=\frac{1}{xz}$ , which implies after multiplying both sides by $yz$ that we get $yz\cdot z^2=1$ and from $y=z^2$ we get $z^5=1\implies z=1\implies x=y=1$ . After checking the solution $(1,1,1)$ on the initial system of equations, we see it works.

Therefore, the answer is $x+y+z=1+1+1=\boxed 3$ .

By AM-GM, $1 = \frac{x + \frac{1}{yz} + z^2}{3} \geq \sqrt[3]{\frac{xz^2}{yz}}$ , i.e. $1 \geq \frac{xz}{y}$ . Similarly, $1 = \frac{y + \frac{1}{xz}}{2} \geq \sqrt{\frac{y}{xz}}$ , i.e. $1\geq \frac{y}{xz}$ . Taking the 2 inequalities above together, we must have $\frac{y}{xz} = 1$ , or $y = xz$ . Substituting this into the 2nd equation and using AM-GM again, $2 = y + \frac{1}{y} \geq 2 \sqrt{y \cdot \frac{1}{y}} = 2.$ Equality must hold in the inequality above, i.e. $y = 1/y$ , or y = 1.

This also implies that xz = 1, or $\frac{1}{z} = x$ . Substituting this into the first equation, we get $x + x + \frac{1}{x^2} = 3, \\ 2x^3 - 3x^2 + 1 = 0, \\ (x-1)^2(2x+1) = 0.$ Since x must be positive, the only solution is x = 1. This implies that the only solution is (1,1,1), so the desired sum is 3.