Rationalize the first term, sqrt(2)sqrt(2)-sqrt(3).

Multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{2}+\sqrt{3}$.

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

Multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{3}-\sqrt{2}$.

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

$$(-2 - \sqrt{6}) - (3 - \sqrt{6})$$ $$= -2 - \sqrt{6} - 3 + \sqrt{6}$$

$$(-2 - 3) + (-\sqrt{6} + \sqrt{6})$$ $$= -5 + 0 = -5$$

The final answer is $-5$.