3. Given $\sqrt{2 \sqrt{3}-3}=\sqrt{\sqrt{3} x}-\sqrt{\sqrt{3} y}(x, y$ are rational numbers). Then $x-y=$ $\qquad$ .
3. 1 .

From the given, $\sqrt{2-\sqrt{3}}=\sqrt{x}-\sqrt{y}$.
Then $\sqrt{x}-\sqrt{y}=\sqrt{\frac{4-2 \sqrt{3}}{2}}=\frac{\sqrt{(\sqrt{3}-1)^{2}}}{\sqrt{2}}$ $=\frac{\sqrt{3}-1}{\sqrt{2}}=\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}$.
Thus, $x+y-2 \sqrt{x y}=\frac{3}{2}+\frac{1}{2}-2 \sqrt{\frac{3}{2} \times \frac{1}{2}}$.
Since $x, y$ are rational numbers, we have $x y=\frac{3}{2} \times \frac{1}{2}, x+y=\frac{3}{2}+\frac{1}{2}$.
Also, $x>y$, so $x=\frac{3}{2}, y=\frac{1}{2} \Rightarrow x-y=1$.