[mrlucky0]

If a benzene is substituted with 2 groups (para with respect to each other; See attachment): a CH₃ and another charged group, say, either Y⁺ or Y^-. In which case would the CH₃ have a "greater" tendency to donate electrons by hyperconjugation? 

I would think that CH₃ would better donate electrons if the substituent was Y⁺ in order to stabilize that positive charge, but my textbook says the donating tendency actually would be greater if the substituent was Y^- rather than Y⁺ can I can't figure out why. Why should CH₃ want to donate electrons if the Y^- was already electron rich? I don't get it.

( I realize that my use "donating tendency" is a little vague, but I had to make my question more specific for the purpose of posting. In case anyone is already familiar, the "donating tendency" is actually assigned a quantitative value. See the table.)

I asked my course TA. He drew an example and explained it to me but I still quite unclear. Here in the two examples, Y^- is O^- and Y⁺ is CH₂^-.

Can anyone convince me that my thinking is wrong?

[Kryolith]

I don't think that a carbanion is more stabilzed by a methyl group than a carbocation. Here are my thoughts:

The stabilization of a carbocation is based on an interaction of the antisymmetric C-H-orbital (which is constructed from the linear combination of two C-H-orbitals) with the p-orbital. The two electrons are in a bonding orbital. Assuming a planar carbanion, it would result in a small destabilization (4 electrons), so this will not occur.

Also supposable would be a form of negative hyperconjugation, which is observed for carbanions substituted with a CF₃ group. The lone pair can be delocalised into the orbital. This stabilizes the system.
Now if you have a CH₃ group instead, the energy of the orbital is too high for an interaction.

==> In your picture the most important resonance structures are the one with the negative charge at the oxygen and the one with the positive charge at the trisubstituted carbon.