9. The highest point of the ellipse obtained by rotating the ellipse $\frac{x^{2}}{2}+y^{2}=1$ counterclockwise by $45^{\circ}$ around the coordinate origin is $\qquad$ .

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9. $\frac{\sqrt{15}}{3}$.

Draw the tangent line to the ellipse at its highest point, then the slope of the tangent line is 0. After rotating the ellipse $45^{\circ}$ clockwise to restore it, the slope of the tangent line is -1.

Let the point of tangency be $\left(x_{0}, y_{0}\right)$, and the tangent line be $\frac{x_{0} x}{2}+y_{0} y=1$, with its slope being $\frac{x_{0}}{2 y_{0}}=-1$. Substituting into the ellipse equation, the point of tangency can be found to be $\left(\frac{2 \sqrt{3}}{3}, \frac{\sqrt{3}}{3}\right)$.
Therefore, the required distance is $\frac{\sqrt{15}}{3}$.