Let a, b be nonzero real numbers. Choose the correct statement.

Option A: 
$$4a^{2} — 3ab + b^{2} \geq 9$$
= $$4a^{2} — 4ab + b^{2} + ab \geq 9$$
= $$\left(2a\ -\ b\right)^2+ab\ \ge\ 9$$

From this we cannot necessarily say that $$ab\ \ge\ 9$$, because we don't know the value of $$\left(2a-b\right)^2$$

If we look at option D which is very similar, 
It says that we have $$ab\ \ge\ 9$$ and we have to prove $$4a^{2} — 3ab + b^{2} \geq 9$$. 
From the inequality we derived above, $$\left(2a\ -\ b\right)^2+ab\ \ge\ 9$$, 
from the option we know that$$ab\ge\ 9$$ , therefore, this term would absolutely be $$\ge\ 9$$ since the minimum value $$\left(2a-b\right)^2$$ can have is zero and even that satisfies the condition. 

Option B is simple calculations, where upon squaring the original statement, we get $$9a^{2} — 12ab + 4b^{2} > 0$$ which is opposite to what are are supposed to check. 

Option C is different, Upon taking a common from the given equation, we get:
$$a\left(9a-12b+4b^2\right)=0$$
Since it's given that a and b are non-zero, the second term must be zero.
Solving this for b we would reach an irrational equation where we cannot solve for b anymore and not reach the value of b= 2a/3
That is, we cannot reach the conclusion that b=2a/3 from the initial equation. 

Therefore, Option D is the correct answer.