6. The integer solutions $(m, n)=$ $\qquad$ for the equation $\frac{1}{m}+\frac{1}{n}-\frac{1}{m n^{2}}=\frac{3}{4}$.
6. $(3,2)$.

Obviously, $m \neq 0, n \neq 0$.
Transform the given equation to
$$
m=\frac{4(n+1)(n-1)}{n(3 n-4)} \text {. }
$$

Since $(n, n+1)=1=(n, n-1)$, it follows that $n \mid 4$.
Thus, $n= \pm 1, \pm 2, \pm 4$.
Upon inspection, only when $n=2$, $m$ is a non-zero integer 3.