Suppose $x>1$ is a real number such that $x+\tfrac 1x = \sqrt{22}$.  What is $x^2-\tfrac1{x^2}$?
1. Given the equation \( x + \frac{1}{x} = \sqrt{22} \), we need to find \( x^2 - \frac{1}{x^2} \).

2. First, square both sides of the given equation:
   \[
   \left( x + \frac{1}{x} \right)^2 = (\sqrt{22})^2
   \]
   \[
   x^2 + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^2} = 22
   \]
   \[
   x^2 + 2 + \frac{1}{x^2} = 22
   \]

3. Subtract 2 from both sides to isolate \( x^2 + \frac{1}{x^2} \):
   \[
   x^2 + \frac{1}{x^2} = 22 - 2
   \]
   \[
   x^2 + \frac{1}{x^2} = 20
   \]

4. Now, we need to find \( x^2 - \frac{1}{x^2} \). Notice that:
   \[
   \left( x - \frac{1}{x} \right)^2 = x^2 - 2 + \frac{1}{x^2}
   \]

5. Let \( y = x - \frac{1}{x} \). Then:
   \[
   y^2 = x^2 - 2 + \frac{1}{x^2}
   \]

6. We already know \( x^2 + \frac{1}{x^2} = 20 \). Therefore:
   \[
   y^2 = 20 - 2
   \]
   \[
   y^2 = 18
   \]

7. Thus:
   \[
   x^2 - \frac{1}{x^2} = \sqrt{18} = 3\sqrt{2}
   \]

The final answer is \( \boxed{18} \)