1. **State the problem:** Given $x=3+\sqrt{8}$, find the value of $\left(x-\frac{1}{x}\right)^2$. 2. **Recall the formula:** We want to evaluate $\left(x-\frac{1}{x}\right)^2 = x^2 - 2 + \frac{1}{x^2}$ using the identity $\left(a-b\right)^2 = a^2 - 2ab + b^2$ and noting $ab=1$ here. 3. **Calculate $x^2$:** $$x^2 = (3+\sqrt{8})^2 = 3^2 + 2 \cdot 3 \cdot \sqrt{8} + (\sqrt{8})^2 = 9 + 6\sqrt{8} + 8 = 17 + 6\sqrt{8}$$ 4. **Calculate $\frac{1}{x}$:** $$\frac{1}{x} = \frac{1}{3+\sqrt{8}} = \frac{3-\sqrt{8}}{(3+\sqrt{8})(3-\sqrt{8})} = \frac{3-\sqrt{8}}{9 - 8} = 3 - \sqrt{8}$$ 5. **Calculate $\frac{1}{x^2}$:** $$\frac{1}{x^2} = (\frac{1}{x})^2 = (3 - \sqrt{8})^2 = 9 - 6\sqrt{8} + 8 = 17 - 6\sqrt{8}$$ 6. **Substitute into the expression:** $$\left(x-\frac{1}{x}\right)^2 = x^2 - 2 + \frac{1}{x^2} = (17 + 6\sqrt{8}) - 2 + (17 - 6\sqrt{8})$$ 7. **Simplify:** $$= 17 + 6\sqrt{8} - 2 + 17 - 6\sqrt{8} = (17 + 17 - 2) + (6\sqrt{8} - 6\sqrt{8}) = 32$$ **Final answer:** $$\boxed{32}$$