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Suppose 1/x = 2/(y + z) = 3/(z + x) = (x^2 - y - z)/(x + y + z).  What is (z - y)/x?

Using the first two equalities  we have that

(y  + z)  =  2x       (1)

And using the second two equalities we have that

2(z + x)  = 3(y + z)  ⇒  2(z + x)  =  3(2x) ⇒   2z = 4x   ⇒  z = 2x   (2)

Sub (1) into the last expression in the equality :

(x^2 - (y + z) )                 x^2 - 2x        x ( x - 2)          x - 2

____________  =        ________   =  _______  =    ____

(x + y + z )                       x + 2x            3x                   3

So  using this and the first expression we have that

x - 2               1

____  =       ____      cross-multiply

   3                 x

x^2 - 2x = 3

x^2 - 2x - 3  =  0     factor

(x - 3) ( x + 1)  =   0

So either x = 3   or    x  = -1

Which means that   z = 6    or  z  = -2

Let x  = -1    and z =  -2

(y + z) = 2x

( y - 2)  = -2

Which means that y  = 0

Note that....when x = -1 , z =  -2    we have that

1/x = 2/(y + z) = 3/(z + x) = (x^2 - y - z)/(x + y + z)

1/ -1  = 2 / (0 - 2)   =  3 / (-2 - 1)   =  [ (-1)^2 - 0 + 2 ] / [ -1 - 0 - 2]  =  

1/-1   =  2/-2  = 3/ -3  =  3/-3

-1  = - 1  = - 1  = - 1

So

(z - y )  / x   =

(-2 - 0) / -1   =

2

BTW......The same result would be found when x = 3  and z  = 6

Suppose \(\dfrac{1}{x} = \dfrac{2}{y + z} = \dfrac{3}{ z + x} = \dfrac{x^2 - y - z}{x + y + z}\).  

What is \(\dfrac{z - y}{x}\) ?

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{1}{x}} &=& \mathbf{\dfrac{3}{ z + x}} \\\\ z + x &=& 3x \\\\ \mathbf{z} &=& \mathbf{2x} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{\dfrac{1}{x}} &=& \mathbf{\dfrac{2}{y + z}} \\\\ y+z &=& 2x \quad | \quad \mathbf{z=2x} \\\\ y+2x &=& 2x \\\\ \mathbf{y} &=& \mathbf{0} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline && \mathbf{\dfrac{z - y}{x}} \quad | \quad \mathbf{y=0} \\\\ &=& \dfrac{z}{x} \quad | \quad \mathbf{z=2x} \\\\ &=& \dfrac{2x}{x} \\\\ &=& \mathbf{2} \\ \hline \end{array}\)