For the following amplifier, why is the output voltage 0 when input voltage is 0?
Whenever you have this type of question, usually the best approach is to compute the base-emitter voltage of the transistors. This is what you tried to do, judging from your comments.

The base-emitter voltage of the NPN transistor is \$V_\mathrm{BB}/2 - v_\mathrm{O}\$. The base-emitter voltage of the PNP transistor is \$-V_\mathrm{BB}/2 - v_\mathrm{O}\$. Applying Kirchhoff’s current law at the output junction and substituting the diode equation yields

$$i_\mathrm{N} - i_\mathrm{P} = I_s e^{V_\mathrm{BB}/(2V_\mathrm{T})} (e^{-v_\mathrm{O}/V_\mathrm{T}} - e^{v_\mathrm{O}/V_\mathrm{T}}) = v_\mathrm{O} / R_\mathrm{L}.$$

The trivial solution of this equation is \$v_\mathrm{O} = 0\$. It is straight forward to show that this is the only solution. Note that the gradient for the LHS is decreasing while the gradient of the RHS is increasing.



Although it is difficult to see, the LHS (blue) has the opposite sign of the RHS (red).