The second and the third fractions suggest substituting $a = \frac{x+1}{y}$, $b = \frac{y}{x-z}$, and $c = \frac{x+y}{z-1}$. Changes are not difficult: $a = \dfrac{x+1}{y} = \dfrac{x}{y} + 1$, so $\frac{x}{y} = a - 1$. $b = \dfrac{y}{x-z} = \dfrac{y+y-x}{x-z} = \frac{c - a}{x-z}$. $c=\dfrac{x+y}{z-1} = \dfrac{x+ a y-a y}{z-1} = \dfrac{x + a \cdot \dfrac{x}{1+a}- a \cdot \dfrac{x}{1+a}} {z-1} = \dfrac{x}{z-1}$. We derived that $a-1 = \frac{c - a}{x-z}$. We substitute the first fraction of $c$ there and simplify: $a-1 = \frac{\dfrac{x}{z-1} - a}{x-z}$. Cross multiplying, $a-1 = \dfrac{x - az -x+ az}{x-z} = \dfrac{\cancel{x} - az \cancel{-x}+ az}{\cancel{x}-z}= a$. Since $y$ and $z$ are also random variable, we have derived that $a-1=a$, $0=1$. Therefore, there is no such $a$ and it is impossible. The numerator of $\frac{x+1}{y}$ is always $1$, so the denominator should be $\pm 1$. When the denominator $ =1$, we know that $y=1$ and $x = 0$ if we substitute into the first and last fraction of the left-hand side of the equation. . When denominator $= -1$, we know that $y=-1$, since variables have to be positive. Putting them into the denominators on the right-hand side of the equation, we get that $x-1 = -z$, so $x = 1 - z$. We know that $z$ has to be in the range $[0,1]$. We substitute $x, y$, and $z$ into the left-hand side of the equation and simplify, \[\dfrac{y}{x-z} = \dfrac{-1}{-z} = \dfrac{1}{z} = \dfrac{1}{1-y}.\] We cross-multiply, $1 - y = z$ (since $z$ is positive), hence $x = 1-z = 1-(1-y) = y$. Since $x = y$, $\frac{x}{y} = \frac{y}{y} = \frac{y}{x-z} = \frac{1}{z} =z = \boxed{ \frac{1}{2}}$.
