1. **State the problem:** We need to find the values of $x$ and $y$ such that $$\frac{(1+i)x-2i}{3+i}+\frac{(2-3i)y+i}{3-i}=i.$$\n\n2. **Rewrite the equation:** The equation involves complex numbers and fractions. We will clear denominators by multiplying both sides by the product $(3+i)(3-i)$. Note that $(3+i)(3-i) = 3^2 - i^2 = 9 - (-1) = 10$.\n\n3. **Multiply both sides by 10:**\n$$10 \left( \frac{(1+i)x-2i}{3+i} + \frac{(2-3i)y+i}{3-i} \right) = 10i.$$\n\n4. **Rewrite each term:**\n$$10 \cdot \frac{(1+i)x-2i}{3+i} = 10 \cdot \frac{(1+i)x-2i}{3+i} = ((1+i)x - 2i)(3 - i)$$\nbecause multiplying numerator and denominator by the conjugate of denominator $3 - i$ cancels denominator. Similarly,\n$$10 \cdot \frac{(2-3i)y + i}{3 - i} = ((2-3i)y + i)(3 + i).$$\n\n5. **Rewrite the equation:**\n$$((1+i)x - 2i)(3 - i) + ((2-3i)y + i)(3 + i) = 10i.$$\n\n6. **Expand the first product:**\n$$(1+i)x(3 - i) - 2i(3 - i) = x(3 + 3i - i - i^2) - 2i(3 - i) = x(3 + 2i + 1) - 2i(3 - i) = x(4 + 2i) - 2i(3 - i).$$\nCalculate $-2i(3 - i)$:\n$$-2i \cdot 3 + 2i \cdot i = -6i + 2i^2 = -6i + 2(-1) = -6i - 2.$$\nSo the first product is:\n$$x(4 + 2i) - 6i - 2 = 4x + 2ix - 6i - 2.$$\n\n7. **Expand the second product:**\n$$((2-3i)y + i)(3 + i) = (2y - 3iy + i)(3 + i) = (2y + i - 3iy)(3 + i).$$\nExpand:\n$$2y(3 + i) + i(3 + i) - 3iy(3 + i) = 6y + 2iy + 3i + i^2 - 9iy - 3i^2 y.$$\nRecall $i^2 = -1$, so:\n$$6y + 2iy + 3i - 1 - 9iy + 3y = (6y + 3y - 1) + (2iy - 9iy + 3i) = (9y - 1) + (-7iy + 3i).$$\n\n8. **Sum both expanded parts:**\n$$(4x + 2ix - 6i - 2) + (9y - 1 - 7iy + 3i) = 10i.$$\nGroup real and imaginary parts:\nReal: $4x - 2 + 9y - 1 = 4x + 9y - 3$\nImaginary: $2ix - 6i - 7iy + 3i = i(2x - 6 - 7y + 3) = i(2x - 7y - 3)$\n\n9. **Rewrite the equation:**\n$$4x + 9y - 3 + i(2x - 7y - 3) = 10i.$$\n\n10. **Equate real and imaginary parts:**\nReal part: $$4x + 9y - 3 = 0$$\nImaginary part: $$2x - 7y - 3 = 10.$$\n\n11. **Simplify imaginary part:**\n$$2x - 7y = 13.$$\n\n12. **Solve the system:**\n$$\begin{cases} 4x + 9y = 3 \\ 2x - 7y = 13 \end{cases}$$\nMultiply second equation by 2:\n$$4x - 14y = 26.$$\nSubtract first equation from this:\n$$(4x - 14y) - (4x + 9y) = 26 - 3 \Rightarrow -23y = 23 \Rightarrow y = -1.$$\n\n13. **Find $x$:**\nSubstitute $y = -1$ into $4x + 9y = 3$:\n$$4x + 9(-1) = 3 \Rightarrow 4x - 9 = 3 \Rightarrow 4x = 12 \Rightarrow x = 3.$$\n\n**Final answer:**\n$$x = 3, \quad y = -1.$$