We know that $\frac{\partial (\hat{a}.\hat{b})}{\partial (\hat{a}.\hat{b})} = 1$
But, if we use distributive law
$\frac{\partial (\hat{a}.\hat{b})}{\partial (\hat{a}.\hat{b})} = \frac{\partial \hat{a}}{\partial (\hat{a}.\hat{b})}.\hat{b} + \hat{a}.\frac{\partial \hat{b}}{\partial (\hat{a}.\hat{b})}$
with $\frac{\partial \hat{a}}{\partial (\hat{a}.\hat{b})} = \frac{1}{b_{x}} \hat{i} + \frac{1}{b_{y}} \hat{j} + \frac{1}{b_{z}} \hat{k}$ and $\frac{\partial \hat{b}}{\partial (\hat{a}.\hat{b})} = \frac{1}{a_{x}} \hat{i} + \frac{1}{a_{y}} \hat{j} + \frac{1}{a_{z}} \hat{k}$
If we use that in the previous equation,
$\frac{\partial (\hat{a}.\hat{b})}{\partial (\hat{a}.\hat{b})} = 3 + 3 = 6$
Where did I go wrong here?
Lets take one dimensional case where, $\hat{a}.\hat{b} = ab$ ;(can be trivially generalized to higher dimensions). Then, what we need to prove is $\frac{d(\hat{a}.\hat{b})}{d(\hat{a}.\hat{b})} = 1$, NOT $\frac{\partial (\hat{a}.\hat{b})}{\partial (\hat{a}.\hat{b})} = 1$.
By distributive law,
$\frac{d(\hat{a}.\hat{b})}{d(\hat{a}.\hat{b})}$ = $\frac{d(\hat{a})}{d(\hat{a}.\hat{b})}.\hat{b}$ + $\hat{a}.\frac{d(\hat{b})}{d(\hat{a}.\hat{b})}$
= $\frac{d(a)}{d(ab)}b + a\frac{d(b)}{d(ab)}$ ...................1
$a = \frac{ab}{b}$
$da = \frac{d(ab)}{b} - \frac{(ab)db}{b^2}$
divide by $d(ab)$,
$\frac{da}{d(ab)} = \frac{1}{b} - \frac{(ab)db}{b^2d(ab)}$ NOT just $\frac{1}{b}$
Substituiting $\frac{da}{d(ab)}$ in equation 1,
$\frac{d(\hat{a}.\hat{b})}{d(\hat{a}.\hat{b})} = 1 - a\frac{db}{d(ab)} + a\frac{db}{d(ab)} = 1$.
Hope this helps.