6. Given that $x$ and $y$ are both negative integers, and satisfy $y=$ $\frac{10 x}{10-x}$. Then the maximum value of $y$ is ( ).
(A) 90
(B) -9
(C) -11
(D) None of the above conclusions are correct
6. D.

From the given conditions, we have
$$
x=\frac{10 y}{y+10}=10-\frac{100}{y+10} \text{. }
$$

For \( x \) to be a negative integer, \( y+10 \) must be a divisor of 100, and \( -10 < y < 0 \).
Therefore, \( y+10=1,2,4,5 \).
Thus, \( y=-9,-8,-6,-5 \).
Substituting and verifying, we find that when \( y=-5 \), \( x=-10 \) satisfies the condition.
Hence, the maximum value of the negative integer \( y \) is -5.