Analytic function from the unit disk to itself with specific values

Question

Let $f:\mathbb{D} \to \mathbb{D}$ be analytic such that $f\left(\frac{1}{4}\right)=\frac{2}{3}$ can $f\left(\frac{1}{3}\right)=-\frac{2}{3}$? Is there a theorem that prevents zeroes or bounds the derivatives

Answer 1

Apply Schwarz-Pick lemma $|\frac{f(\frac{1}{3})-f(\frac{1}{4})}{1-\bar{f(\frac{1}{3})}f(\frac{1}{4})}|\le |\frac{\frac{1}{3}-\frac{1}{4}}{1-\frac{1}{3}\frac{1}{4}}|$

So we need $\frac{\frac{4}{3}}{\frac{13}{9}} \le \frac{1}{11}$ and that is incorrect, so no such $f$ exists