6. Given that $x, y$ are natural numbers, and $x^{2}+y^{2}+6$ is a multiple of $x y$. Then $\frac{x^{2}+y^{2}+6}{x y}$ is a ( ).
(A) Perfect square
(B) Perfect cube
(C) Perfect fourth power
(D) Prime number
6. (B).

Given $x y \mid x^{2}+y^{2}+6$, we can set $k=\frac{x^{2}+y^{2}+6}{x y}$. Then we have $x^{2}-k y \cdot x+y^{2}+6=0$.

This is a quadratic equation in $x$, with its two roots denoted as $x_{1}$ and $x_{2}$. Thus, $\left\{\begin{array}{l}x_{1}+x_{2}=k y, \\ x_{1} \cdot x_{2}=y^{2}+6 .\end{array}\right.$
Therefore, $x_{1}, x_{2}$ are both positive integers.
In (1), if $x=y$, then (1) simplifies to $(2-k) x^{2}+6=0$. Hence, $x^{2} \mid 6$, which must have $x=y=1$.
Assume $x_{1}>y$, then when $y \geqslant 3$,
$$
x_{2}=\frac{\dot{y}^{2}+6}{x_{1}} \leqslant \frac{y^{2}+6}{y+1}=y+1+\frac{5-2 y}{y+1}<y+1 \text {. }
$$

Thus, $x_{2} \leqslant y<x_{1}$.
When $y \doteq 2$, from $x_{1} \mid y^{2}+6$ we know $x_{1} \geqslant 4$. Therefore, $x_{2}=\frac{y^{2}+6}{x_{1}}=\frac{10}{x_{1}}<x$.
From the above, if $x_{1} \geqslant y \geqslant 2$, then we can find another solution $(y, x_{2})$, satisfying $y \geqslant x_{2}$, and $y+x_{2}<x_{1}+y$. This process cannot continue indefinitely, so we must eventually obtain a solution $(x, 1)$. Since $k$ is a positive integer, we have $x!1^{2} .+6$, which means $x=1$ or 7. Thus, the corresponding $y$ is 7 or 1, i.e., $\frac{x^{2}+y^{2}+6}{x y}=8$ is a perfect cube.