Find the balanced redox reaction of Bi3+ + OH- + MnO4                MnO2 + BiO3-?

"MnO₄" does not exist. It should be "MnO₄⁻" instead.

Oxidation half equation: Bi³⁺ + 6OH⁻ → BiO₃⁻ + 3H₂O + 2e⁻
Reduction half equation: MnO₄⁻ + 2H₂O + 3e → MnO₂ + 4OH⁻

(Oxidation half equation)×3 + (Reduction half equation)×2, and cancel 6e⁻, 8OH⁻ and 4H₂O on the both sides. The balanced overall redox equation is:
3Bi³⁺ + 10OH⁻ + 2MnO₄⁻ → 3BiO₃⁻ + 2MnO₂ + 5H₂O

Half-reactions:

Bi^3+ ---> BiO3^-
MnO4^- ---> MnO2

Balance in acidic solution:

3H2O + Bi^3+ ---> BiO3^- + 6H^+ + 2e^-
3e^- + 4H^+ + MnO4^- ---> MnO2 + 2H2O

Equalize electrons:

9H2O + 3Bi^3+ ---> 3BiO3^- + 18H^+ + 6e^-
6e^- + 8H^+ + 2MnO4^- ---> 2MnO2 + 4H2O

Add and eliminate like items:

5H2O + 3Bi^3+ + 2MnO4^- ---> 3BiO3^- + 2MnO2 + 10H^+

Change to basic solution:

10OH^- + 5H2O + 3Bi^3+ + 2MnO4^- ---> 3BiO3^- + 2MnO2 + 10H2O

Remove water:

10OH^- + 3Bi^3+ + 2MnO4^- ---> 3BiO3^- + 2MnO2 + 5H2O

More examples:

https://www.chemteam.info/Redox/Balance-Redox-Base.html