Let $f:(0,\infty) \rightarrow (0,\infty)$ be a function such that
\[ 
10\cdot \frac{x+y}{xy}=f(x)\cdot f(y)-f(xy)-90
\]
for every $x,y \in (0,\infty)$. What is $f(\frac 1{11})$?

$ 
\textbf{(A)}\ 1
\qquad\textbf{(B)}\ 11
\qquad\textbf{(C)}\ 21
\qquad\textbf{(D)}\ 31
\qquad\textbf{(E)}\ \text{There is more than one solution}
$
1. Start by plugging in \( y = 1 \) into the given functional equation:
   \[
   10 \cdot \frac{x + 1}{x \cdot 1} = f(x) \cdot f(1) - f(x) - 90
   \]
   Simplifying the left-hand side:
   \[
   10 \cdot \left( \frac{x + 1}{x} \right) = 10 \left( 1 + \frac{1}{x} \right) = 10 + \frac{10}{x}
   \]
   Thus, the equation becomes:
   \[
   10 + \frac{10}{x} = f(x) \cdot f(1) - f(x) - 90
   \]

2. Rearrange the equation to isolate \( f(x) \):
   \[
   10 + \frac{10}{x} + 90 = f(x) \cdot f(1) - f(x)
   \]
   \[
   100 + \frac{10}{x} = f(x) \cdot f(1) - f(x)
   \]
   \[
   100 + \frac{10}{x} = f(x) (f(1) - 1)
   \]

3. Now, plug in \( x = 1 \) and \( y = 1 \) into the original equation:
   \[
   10 \cdot \frac{1 + 1}{1 \cdot 1} = f(1) \cdot f(1) - f(1) - 90
   \]
   Simplifying the left-hand side:
   \[
   10 \cdot 2 = 20
   \]
   Thus, the equation becomes:
   \[
   20 = f(1)^2 - f(1) - 90
   \]
   \[
   f(1)^2 - f(1) - 110 = 0
   \]

4. Solve the quadratic equation for \( f(1) \):
   \[
   f(1) = \frac{1 \pm \sqrt{1 + 4 \cdot 110}}{2} = \frac{1 \pm \sqrt{441}}{2} = \frac{1 \pm 21}{2}
   \]
   \[
   f(1) = 11 \quad \text{or} \quad f(1) = -10
   \]
   Since \( f \) maps \((0, \infty)\) to \((0, \infty)\), \( f(1) = -10 \) is not valid. Therefore, \( f(1) = 11 \).

5. Substitute \( f(1) = 11 \) back into the equation:
   \[
   100 + \frac{10}{x} = f(x) \cdot 10
   \]
   \[
   f(x) = \frac{100 + \frac{10}{x}}{10} = 10 + \frac{1}{x}
   \]

6. Finally, find \( f\left( \frac{1}{11} \right) \):
   \[
   f\left( \frac{1}{11} \right) = 10 + 11 = 21
   \]

The final answer is \( \boxed{21} \).