If I try this question in different ways I get a different answer:

For positive $\displaystyle x,y,z$,

$\displaystyle \frac{y}{x-z}=\frac{x+y}{z}=\frac{x}{y}$

Find the numerical value of $\displaystyle \frac{x}{y}$

(Using the fact that $\displaystyle \frac{a}{b}=\frac{c}{d}=\frac{a+kc}{b+kd}$),

Solution 1:

$\displaystyle \frac{y+(x+y)}{x-z+(z)}=\frac{x}{y}$

$\displaystyle \implies \frac{2y+x}{x}=\frac{x}{y}$

$\displaystyle \implies 2y+x=y$

$\displaystyle \frac{x}{y}=-1$

Solution 2:

$\displaystyle \frac{x}{y}=\frac{y+(x+y)+(x)}{x-z+(z)+(y)}=\frac{2(x+y)}{x+y}=2$