6 If real numbers $x, y$ satisfy $(x-3)^{2}+4(y-1)^{2}=4$, then the maximum and minimum values of $\frac{x+y-3}{x-y+1}$ are ( ).
(A) 1,0
(B) $0,-1$
(C) $1,-1$
(D) $\frac{1}{2},-\frac{1}{2}$
6 Let
$$
\left\{\begin{array}{l}
x=3+2 \cos \theta \\
y=1+\sin \theta
\end{array}\right.
$$

then
$$
t=\frac{x+y-3}{x-y+1}=\frac{2 \cos \theta+\sin \theta+1}{2 \cos \theta-\sin \theta+3},
$$

which simplifies to
$$
(2 t-2) \cos \theta-(t+1) \sin \theta+3 t-1=0,
$$

Using the universal formula, we get
$$
(t+1) \tan ^{2} \frac{\theta}{2}-2(t+1) \tan \frac{\theta}{2}+5 t-3=0 .
$$

When $t \neq-1$,
$$
\frac{1}{4} \Delta=(t+1)^{2}-(t+1)(5 t-3) \geqslant 0,
$$

Thus, $-1<t \leqslant 1$; when $t=-1$, solving yields $\cos \theta=-1$, which is valid. Therefore, the answer is C.