$f(\theta)$ turns out to be this
$sin(\theta)\begin{vmatrix} \frac{\sqrt 3}{2} & \frac{1}{\sqrt 3} \\ \frac{1}{2}&\sqrt 3 \end{vmatrix}-cos(\theta)\begin{vmatrix} \frac{1}{2} &\frac{1}{\sqrt 3} \\ \frac{\sqrt 3}{2} &\sqrt 3 \end{vmatrix}+tan(\theta)\begin{vmatrix} \frac{1}{2} &\frac{\sqrt 3}{2} \\ \frac{\sqrt 3}{2} &\frac{1}{2} \end{vmatrix}$
Now, i can rewrite the above equation as
$f(\theta)=asin(\theta)-bcos(\theta)+ctan(\theta)$
for some non-zero constants $a,b,c$
$f'(\theta)=acos(\theta)+bsin(\theta)+csec^2(\theta)$
now at $\theta=\frac{\pi}{4} \in (\frac{\pi}{6},\frac{\pi}{3})$, $f'(\theta)\neq0$
So, II clause is true.
Please let me know if somewhere my this claim for II is incorrect.