Find the sum of all possible values of $f(2)$ such that
$f(x)f(y)-f(xy) = \frac{y}{x}+\frac{x}{y}$, for every positive real numbers $x,y$

$ \textbf{(A)}\ \frac{5}{2}
\qquad\textbf{(B)}\ -\frac{5}{4}
\qquad\textbf{(C)}\ \frac{5}{4}
\qquad\textbf{(D)}\ \frac{3}{2}
\qquad\textbf{(E)}\ \text{None}
$
1. Start by substituting \( x = y = 1 \) into the given functional equation:
   \[
   f(x)f(y) - f(xy) = \frac{y}{x} + \frac{x}{y}
   \]
   This simplifies to:
   \[
   f(1)f(1) - f(1) = \frac{1}{1} + \frac{1}{1}
   \]
   \[
   f(1)^2 - f(1) = 2
   \]
   This is a quadratic equation in \( f(1) \):
   \[
   f(1)^2 - f(1) - 2 = 0
   \]
   Solving this quadratic equation using the quadratic formula \( f(1) = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1 \), \( b = -1 \), and \( c = -2 \):
   \[
   f(1) = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm 3}{2}
   \]
   Thus, the solutions are:
   \[
   f(1) = 2 \quad \text{or} \quad f(1) = -1
   \]

2. Next, substitute \( x = 2 \) and \( y = 1 \) into the original equation:
   \[
   f(2)f(1) - f(2) = \frac{1}{2} + \frac{2}{1}
   \]
   Simplifying the right-hand side:
   \[
   f(2)f(1) - f(2) = \frac{1}{2} + 2 = \frac{5}{2}
   \]
   This can be factored as:
   \[
   f(2)(f(1) - 1) = \frac{5}{2}
   \]

3. Consider the case when \( f(1) = 2 \):
   \[
   f(2)(2 - 1) = \frac{5}{2}
   \]
   \[
   f(2) = \frac{5}{2}
   \]

4. Consider the case when \( f(1) = -1 \):
   \[
   f(2)(-1 - 1) = \frac{5}{2}
   \]
   \[
   f(2)(-2) = \frac{5}{2}
   \]
   \[
   f(2) = -\frac{5}{4}
   \]

5. The sum of all possible values of \( f(2) \) is:
   \[
   \frac{5}{2} + \left(-\frac{5}{4}\right) = \frac{5}{2} - \frac{5}{4} = \frac{10}{4} - \frac{5}{4} = \frac{5}{4}
   \]

The final answer is \(\boxed{\frac{5}{4}}\).