I'm trying to find positive integer solutions to the following diophantine equation:
$$y = \frac{x z}{-x - z + x z}$$
The first thing I did was split the fraction as follows:
$$ \frac{x z}{-x - z + x z} = 1 + \frac{x+z}{x z-x-z}$$
This doesn't seem to help any though. The only thing I can pin down is that the numerator of the fraction must be larger than the denominator to get a positive value for y, placing a constraint on $x$ and $z$ as follows:
$$z>2\land x<\frac{2 z}{z-2}$$.
Now what though?
If $-x-z+xz=0$, then $(x-1)(z-1)=1$, so $(x,z)=(2,2)$, which doesn't give a solution.
If $-x-z+xz\neq 0$, then the equation is equivalent to $xy+yz+zx=xyz$, i.e. $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1$.
Let wlog $x\ge y\ge z\ge 1$. Then $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\le \frac{3}{z}$, so $z\le 3$.
If $z=1$, then $\frac{1}{x}+\frac{1}{y}=0$; no solutions.
If $z=2$, then $\frac{1}{x}+\frac{1}{y}=\frac{1}{2}$, i.e. $2y+2x=xy$, i.e. $(x-2)(y-2)=4$.
$x-2\ge y-2\ge -1$, so either $(x-2,y-2)=(4,1)$ or $(2,2)$. We get the solutions $(x,y,z)=(6,3,2),(4,4,2)$.
If $z=3$, then $\frac{1}{x}+\frac{1}{y}=\frac{2}{3}$, so $3y+3x=2xy$, so $(2x-3)(2y-3)=9$.
$2x-3\ge 2y-3\ge -1$, so either $(2x-3,2y-3)=(9,1)$ or $(3,3)$. In the first case $y=2<z=3$. In the second case we get the solution $(x,y,z)=(3,3,3)$.
All the solutions are $$(x,y,z)=(6,3,2),(6,2,3),(3,6,2),(3,2,6),(2,6,3),\\(2,3,6),(4,4,2),(4,2,4),(2,4,4),(3,3,3)$$