Let $ABCD$ be a square and $X$ a point such that $A$ and $X$ are on opposite sides of $CD$. The lines $AX$ and $BX$ intersect $CD$ in $Y$ and $Z$ respectively. If the area of $ABCD$ is $1$ and the area of $XYZ$ is $\frac{2}{3}$, determine the length of $YZ$
1. Let $ABCD$ be a square with side length $s$. Given that the area of $ABCD$ is $1$, we have:
   \[
   s^2 = 1 \implies s = 1
   \]

2. Let $X$ be a point such that $A$ and $X$ are on opposite sides of $CD$. The lines $AX$ and $BX$ intersect $CD$ at points $Y$ and $Z$ respectively.

3. Given that the area of $\triangle XYZ$ is $\frac{2}{3}$, we need to determine the length of $YZ$.

4. Let $YZ = x$ and the altitude from $X$ to $YZ$ be $h$. The area of $\triangle XYZ$ can be expressed as:
   \[
   \frac{1}{2} \cdot x \cdot h = \frac{2}{3}
   \]
   Solving for $xh$, we get:
   \[
   xh = \frac{4}{3}
   \]

5. Since $AB = 1$, by similar triangles $\triangle XYZ$ and $\triangle XAB$, we have:
   \[
   \frac{x}{1} = \frac{h}{1 + h}
   \]
   Simplifying, we get:
   \[
   x = \frac{h}{1 + h}
   \]

6. We know that $x = \frac{4}{3h}$ from the area equation. Substituting this into the similarity ratio, we get:
   \[
   \frac{4}{3h} = \frac{h}{1 + h}
   \]

7. Cross-multiplying to solve for $h$, we get:
   \[
   4(1 + h) = 3h^2
   \]
   Simplifying, we get:
   \[
   4 + 4h = 3h^2
   \]
   Rearranging, we get:
   \[
   3h^2 - 4h - 4 = 0
   \]

8. Solving the quadratic equation $3h^2 - 4h - 4 = 0$ using the quadratic formula $h = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 3$, $b = -4$, and $c = -4$, we get:
   \[
   h = \frac{4 \pm \sqrt{16 + 48}}{6} = \frac{4 \pm \sqrt{64}}{6} = \frac{4 \pm 8}{6}
   \]
   This gives us two solutions:
   \[
   h = 2 \quad \text{or} \quad h = -\frac{2}{3}
   \]
   Since $h$ must be positive, we have:
   \[
   h = 2
   \]

9. Substituting $h = 2$ back into the equation $x = \frac{4}{3h}$, we get:
   \[
   x = \frac{4}{3 \cdot 2} = \frac{4}{6} = \frac{2}{3}
   \]

The final answer is $\boxed{\frac{2}{3}}$