Simplify the expressions for x and y.

We are given $x = \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}$ and $y = \frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}-\sqrt{3}}$.
Notice that $y = \frac{1}{x}$.
First, rationalize the denominator for $x$:

$$x = \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}} \times \frac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}$$ $$x = \frac{(\sqrt{2}-\sqrt{3})^2}{(\sqrt{2})^2 - (\sqrt{3})^2}$$ $$x = \frac{2 - 2\sqrt{2}\sqrt{3} + 3}{2 - 3}$$ $$x = \frac{5 - 2\sqrt{6}}{-1}$$ $$x = -5 + 2\sqrt{6}$$

Since $y = \frac{1}{x}$, we can also find $y$:

$$y = \frac{1}{-5 + 2\sqrt{6}} \times \frac{-5 - 2\sqrt{6}}{-5 - 2\sqrt{6}}$$ $$y = \frac{-5 - 2\sqrt{6}}{(-5)^2 - (2\sqrt{6})^2}$$ $$y = \frac{-5 - 2\sqrt{6}}{25 - (4 \times 6)}$$ $$y = \frac{-5 - 2\sqrt{6}}{25 - 24}$$ $$y = -5 - 2\sqrt{6}$$

Since $y = \frac{1}{x}$, their product is:

$$xy = x \cdot \frac{1}{x} = 1$$

$$x^2 = (-5 + 2\sqrt{6})^2$$ $$x^2 = (-5)^2 + 2(-5)(2\sqrt{6}) + (2\sqrt{6})^2$$ $$x^2 = 25 - 20\sqrt{6} + 4 \times 6$$ $$x^2 = 25 - 20\sqrt{6} + 24$$ $$x^2 = 49 - 20\sqrt{6}$$ $$y^2 = (-5 - 2\sqrt{6})^2$$ $$y^2 = (-5)^2 + 2(-5)(-2\sqrt{6}) + (-2\sqrt{6})^2$$ $$y^2 = 25 + 20\sqrt{6} + 4 \times 6$$ $$y^2 = 25 + 20\sqrt{6} + 24$$ $$y^2 = 49 + 20\sqrt{6}$$

$$3x^2 - 5xy + 3y^2 = 3(49 - 20\sqrt{6}) - 5(1) + 3(49 + 20\sqrt{6})$$ $$= 147 - 60\sqrt{6} - 5 + 147 + 60\sqrt{6}$$ $$= (147 + 147 - 5) + (-60\sqrt{6} + 60\sqrt{6})$$ $$= 294 - 5 + 0$$ $$= 289$$

The value of $3x^2 - 5xy + 3y^2$ is $\boxed{\text{289}}$.