Simplification Helps

Changing the given to common denominator reveals that $\frac{x}{(x-y)(x-z)} + \frac{y}{(y-x)(y-z)} + \frac{z}{(z-x)(z-y)}=$ $\frac{x(y-z)+y(z-x)+z(x-y)}{(x-y)(x-z)(y-z)}=$ $\frac{xy-xz+yz-xy+xz-yz}{(x-y)(x-z)(y-z)} =$ $\frac{0}{(x-y)(x-z)(y-z)} = 0.$ And $0$ is not a positive integer.

$\Rightarrow$ There exist exactly zero ordered tuples of distinct positive integers $(x,y,z)$ satisfying $\frac{x}{(x-y)(x-z)} + \frac{y}{(y-x)(y-z)} + \frac{z}{(z-x)(z-y)} = k$ where $k$ is a positive integer.