6 Find all pairs of positive integers $(x, y)$, such that $x^{y}=y^{x-y}$.

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If $x=1$, then $y=1$; if $y=1$, then $x=1$.
If $x=y$, then $x^{y}=1$, so $x=y=1$.
Now consider the case $x>y \geqslant 2$. By the problem statement,
$$
12 y \text {, and } y \mid x \text {. }
$$

Let $x=k y$, then $k \geqslant 3$, and
$$
\begin{array}{c}
k^{y}=y^{(k-2) y}, \\
k=y^{k-2} .
\end{array}
$$

Therefore,
since $y \geqslant 2$, we have $y^{k-2} \geqslant 2^{k-2}$, and it is easy to prove by mathematical induction or the binomial theorem that when $k \geqslant 5$, $2^{k}>4 k$, so $k$ can only be 3 or 4.
When $k=3$, $y=3, x=9$; when $k=4$, $y=2, x=8$.
Thus, all the required pairs of positive integers are $(1,1),(9,3),(8,2)$.