6. In the Cartesian coordinate system, the equation
$$
\frac{|x+y|}{2 a}+\frac{|x-y|}{2 b}=1
$$
( $a, b$ are two different positive numbers) represents the curve ( ).
(A) Triangle
(B) Square
(C) Non-square rectangle
(D) Non-square rhombus
6. (D)

Method one: The lines $l_{1}: x-y=0$ and $l_{2}: x+y=0$ divide the plane $X O Y$ into 4 regions:
( I) $\left\{\begin{array}{l}x+y \geqslant 0, \\ x-y \geqslant 0,\end{array}\right.$
( I ) $\left\{\begin{array}{l}x+y \geqslant 0, \\ x-y \leqslant 0,\end{array}\right.$
(1) $\left\{\begin{array}{l}x+y \leqslant 0, \\ x-y \leqslant 0,\end{array}\right.$
(N) $\left\{\begin{array}{l}x+y \leqslant 0, \\ x-y \geqslant 0 .\end{array}\right.$
In region (I), the equation
$$
\frac{|x+y|}{2 a}+\frac{|x-y|}{2 b}=1
$$

becomes
$$
\frac{x+y}{2 a}+\frac{x-y}{2 b}=1 \text {. }
$$

It represents the line $l_{3}$. Let $l_{3}$ intersect $l_{1}$ at point $A$. Then
$$
\left\{\begin{array}{l}
\frac{x+y}{2 a}+\frac{x-y}{2 b}=1, \\
x-y=0
\end{array}\right.
$$

gives the coordinates of point $A$ as $(a, a)$.
Let $l_{3}$ intersect $l_{2}$ at point $B$, then
$$
\left\{\begin{array}{l}
\frac{x+y}{2 a}+\frac{x-y}{2 b}=1, \\
x+y=0
\end{array}\right.
$$

gives the coordinates of point $B$ as $(b,-b)$.
Therefore, the curve represented by (1) in region (I) is the line segment $A B$.

Similarly, in region (1), equation (1) becomes $\frac{x+y}{2 a}+\frac{y-x}{2 b}=1$, representing the line $l_{4}$. $l_{4}$ intersects $l_{1}$ at point $A$, and $l_{4}$ intersects $l_{2}$ at point $D$, forming the line segment $A D$. The curve represented by (1) in region (1) is the line segment $C D$, and the curve represented by (1) in region (N) is the line segment $C B$.

Since $A O=C O, B O=D O, A C \neq B D, A C \perp B D$, the quadrilateral $A B C D$ represented by (1) is a non-square rhombus. Therefore, the answer is (D).

Method two: Rotate the Cartesian coordinate system $X O Y$ counterclockwise by $45^{\circ}$ to get the new coordinate system $X^{\prime} O Y^{\prime}$. The coordinates of point $P$ in the coordinate system $X^{\prime} O Y^{\prime}$ are $\left(x^{\prime}, y^{\prime}\right)$, and in the coordinate system $X O Y$ are $(x, y)$. Then
$$
\left\{\begin{array}{l}
x^{\prime}=\frac{1}{\sqrt{2}}(x+y), \\
y^{\prime}=\frac{1}{\sqrt{2}}(-x+y) .
\end{array}\right.
$$

The equation $\frac{|x+y|}{2 a}+\frac{|x-y|}{2 b}=1$ in the problem
transforms into $\frac{\left|x^{\prime}\right|}{a}+\frac{\left|y^{\prime}\right|}{b}=\sqrt{2}$.
Clearly, the curve represented by (2) is symmetric about the $X^{\prime}$-axis and the $Y^{\prime}$-axis. In the first quadrant of $X^{\prime} O Y^{\prime}$, (2) becomes $\frac{x^{\prime}}{a}+\frac{y^{\prime}}{b}=\sqrt{2}$, which is the line segment $A B$, where $A(\sqrt{2} a, 0), B(0, \sqrt{2} b)$.

By symmetry, in the first quadrant, equation (2) is the line segment $B C$, where $C(-\sqrt{2} a, 0)$,

in the third quadrant, equation (2) is the line segment $C D$, where $D(0, -\sqrt{2} b)$,
and in the fourth quadrant, equation (2) is the line segment $A D$.
By symmetry, $A B=B C=C D=A D$. Since $a \neq b$, $A C \neq B D$. Therefore, $A B C D$ is a non-square rhombus. Hence, the answer is (D).