867. Solve the equation in natural numbers

$$
x+\frac{y}{z}=\frac{x^{2} y}{z}
$$
$\triangle$ Transform the equation:

$$
x z+y=x^{2} y, \quad x z=y\left(x^{2}-1\right)
$$

From here, the product $y\left(x^{2}-1\right)$ is divisible by $x$. But the numbers $x$ and $x^{2}-1$ are coprime (think about how to prove this, using the solutions of problems 454 and 466 from § 11 as examples). Then $y$ is divisible by $x: y=k x$, where $k$ is a natural number. We get:

$$
x z=k x\left(x^{2}-1\right), \quad z=k\left(x^{2}-1\right)
$$

In the last equation, $x$ must be greater than 1.

Answer: $\left(x ; k x ; k\left(x^{2}-1\right)\right)$, where $k \in N, x \in N, x>1$