Chọn phương án D.

Điều kiện: \(\begin{cases}
0< x\neq1\\
0< y\neq1\\
x-y>0
\end{cases}\Leftrightarrow0< y< x\neq1\).

Ta có$$\begin{aligned}
\log_xy=\log_yx\Leftrightarrow&\log_xy=\dfrac{1}{\log_xy}\\
\Leftrightarrow&\left(\log_xy\right)^2=1\\
\Leftrightarrow&\left[\begin{array}{l}\log_xy=1\\ \log_xy=-1\end{array}\right.\\
\Leftrightarrow&\left[\begin{array}{ll}x=y &\text{(loại)}\\ y=x^{-1}=\dfrac{1}{x}\end{array}\right.
\end{aligned}$$
Ta lại có$$\begin{aligned}
&\,\log_x(x-y)=\log_y(x+y)\\
\Leftrightarrow&\,\log_x\left(x-\dfrac{1}{x}\right)=\log_{\tfrac{1}{x}}\left(x+\dfrac{1}{x}\right)\\
\Leftrightarrow&\,\log_x\dfrac{x^2-1}{x}=\log_{x^{-1}}\dfrac{x^2+1}{x}\\
\Leftrightarrow&\,\log_x\dfrac{x^2-1}{x}=-\log_x\dfrac{x^2+1}{x}\\
\Leftrightarrow&\,\log_x\dfrac{x^2-1}{x}=\log_x\left(\dfrac{x^2+1}{x}\right)^{-1}\\
\Leftrightarrow&\,\log_x\dfrac{x^2-1}{x}=\log_x\dfrac{x}{x^2+1}\\
\Leftrightarrow&\,\dfrac{x^2-1}{x}=\dfrac{x}{x^2+1}\\
\Leftrightarrow&\,x^4-1=x^2\\
\Leftrightarrow&\,x^4-x^2-1=0\\
\Leftrightarrow&x^2=\dfrac{1+\sqrt{5}}{2}
\end{aligned}$$
Khi đó:

• \(xy=x\cdot\dfrac{1}{x}=1\)
• \(y^2=\dfrac{1}{x^2}=\dfrac{2}{1+\sqrt{5}}\)

Vậy \(x^2+xy-y^2=2\).