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Put $\alpha =-\log |c|>0$. Since $f$ has no zeros and the unit disc is simply connected, it has a holomorphic logarithm. Thus $F=-\log f$ is holomorphic with $\mathrm{Re}F>0$ and $\mathrm{Re}F(0)=\alpha$. The Cayley transform between the half-plane and disk givesThe Schwarz lemma yields $|{\varphi }^{\prime }(0)|\le 1$, so the Carathéodory derivative bound for the right half-plane gives $|F^{\prime }(0)|\le 2\alpha$. Since $f^{\prime }=-F^{\prime }f$, the sharp answer isEquality in the Schwarz lemma holds exactly when $\varphi (z)=\omega z$, $|\omega |=1$. Solving for $F$ and exponentiating gives exactly the extremizersTheir logarithmic real part is $\alpha \mathrm{Re}[(1+\omega z)/(1-\omega z)]>0$, so each is indeed a zero-free map into the unit disc with the prescribed value at zero.

For a fixed $w\in \mathbb{D}$, put $r=|w|$. Apply the Harnack inequality on the unit disk to the positive harmonic function $\mathrm{Re}F$ and then exponentiate its negative. The zero-free Schwarz lemma givesFor $w\ne 0$ both endpoints are attained: take $\omega w=r$ for the lower endpoint and $\omega w=-r$ for the upper endpoint in the displayed extremizers. At zero both are $|c|$. If the question is interpreted as allowing $w$ to vary as well, the sharp infimum and supremum over all admissible pairs $(f,w)$ are respectively $0$ and $1$, approached as $r\uparrow 1$; neither value is attained inside the unit disc. An individual function need not have these global extremal values: the constant function $f=c$ is a simple example.