Here are the steps I took:
i) Found a common denominator on the right side to add the fractions
ii) Took the reciprocals of both sides
iii) Multiplied top and bottom by the conjugate to get 'i' out of the bottom.
Given problem: 1/z = (2/(2+3i)) + (1/(3-2i))
i) 1/z = (2/(2+3i))*[(3-2i)/(3-2i)] + (1/(3-2i))*[(2+3i)/(2+3i)]
= [(6-4i)+(2+3i)] / (6+5i-6i2)
→ 1/z = (8-i) / (12+5i) *Note i2 = -1
ii) z = (12+5i) / (8-i)
iii) z = [(12+5i) / (8-i)] * [(8+i)/(8+i)]
= (96+52i+5i2) / (64-i2)
= (91+52i) / 65
= (7+4i) / 5
= 7/5 +(4/5)i
Hope the above helps, and thank you for posting the question. If you have any questions about any of the steps, please let me know.