The value of the logarithmic expression can never be $\ldots$ The value of the logarithmic expression  $\log_x \dfrac{x}{y} +\log_y \dfrac{y}{x},\text{where}\quad x\geq y>1\quad$  can never be 
$\bf\text{options}$ a.) $-1\quad$   b.)$\quad0.5\quad$ c.) $\quad0\quad$ d.) $\quad1$
$$\begin{align}\log_x \dfrac{x}{y} +\log_y \dfrac{y}{x}&=1-\log_x y +1-\log_y x\\
&=\frac{2\log_x y}{\log_x y}-\frac{(\log_x y)^2}{\log_x y} -\frac{1}{\log_x y}\\
&=-\frac{ (\log _{x}y-1)^{2} }{ \log _{x}y}\end{align}$$
here if i input $x=y=2$ then option $0$ is removed , not sure how to deduce the rest.
Wolfram does gives the answer as option $1$

Note: We start in the same way  as OP did and then we  proceed accordingly.

We consider
\begin{align*}
\log_x\frac{x}{y}+\log_y\frac{y}{x}\qquad\qquad\qquad x\geq y>1\tag{1}
\end{align*}
  
  
*
  
*Validity of domain:
  At first, we check the domain and observe that since $\log_bz$ is defined for any $z>0$ and for $b>0, b\ne 1$ the expression (1) is valid for all $x\geq y>1$.
  
*Since $\log_bx=\frac{\log_kx}{\log_kb}$  for any bases $b,k>1$ we obtain
  \begin{align*}
\log_xy=\frac{\log_yy}{\log_yx}=\frac{1}{\log_yx}\tag{2}
\end{align*}
  
*We also keep in mind that for $x\geq y > 1$
  \begin{align*}
0<\log_xy\leq 1 \tag{3}
\end{align*}

Options b.) and d.)

We start now similarly as OP did:
  \begin{align*}
\log_x&\frac{x}{y}+\log_y\frac{y}{x}\\
&=\log_xx-\log_xy+\log_yy-\log_yx\\
&=2-\log_xy-\frac{1}{\log_xy}\\
&=-\frac{1}{\log_xy}\left((\log_xy)^2-2\log_xy+1\right)\\
&=-\frac{\left(\log_xy-1\right)^2}{\log_xy}\leq 0\tag{4}
\end{align*}
In (4) we use $0<\log_xy\leq 1$ accordingly to (3).
Conclusion: The expression (1) is less or equal to zero whenever $x\geq y >1$. Therefore it can never be option b.) with value $0.5$ or option d.) with value $1$

Options a.) and c.)

We now set $z=\log_xy$ and consider accordingly to (4) the expression
\begin{align*}
-\frac{\left(z-1\right)^2}{z}=a\qquad\qquad a \in \mathbb{R}
\end{align*}
  or equivalently
  \begin{align*}
z^2-(a-2)z+1=0
\end{align*}
  with solutions
  \begin{align*}
z_{1,2}=1-\frac{a}{2}\pm\frac{1}{2}\sqrt{a(a-4)}
\end{align*}
  Now, if we set accordingly to option c.) $a=0$ we see
  $z_{1,2}=1$
  and therefore
  $$\log_xy=1 \Rightarrow x=y$$
Conclusion: Letting $x=y>1$, we get 
\begin{align*}
\log_x&\frac{x}{y}+\log_y\frac{y}{x}\\
&=\log_x\frac{x}{x}+\log_x\frac{x}{x}\\
&=2\log 1\\
&=0
\end{align*}
  and so we can't skip option c.) 

Next we set $a=-1$ and get 
  $z_{1,2}=\frac{1}{2}(3\pm \sqrt{5})$
Conclusion: Since $z_2=\frac{1}{2}(3-\sqrt{5})< 1$ we see that according to (3) we also find a solution for option a.) with value $-1$

