$|f(z)|\leq 1$, then possible value of $(e^f)''(0)$ 
Let $f$ be an analytic function on $\bar{D}=\{z\in \mathbb{C}:|z|\leq1\}$. Assume that $|f(z)|\leq1~ \forall z\in \bar{D}$. Then, which of the following is NOT a possible value of $(e^f)''(0)$.

(i)6 (ii)2 (iii)$\frac{7}{9}e^{\frac{1}{9}}$ (iv)$\sqrt{2}+i\sqrt{2}$.
I calculated $(e^f)'=f'e^f$. Here if $(e^f)'(0)$ were $0$ then I could have used Schwarz Lemma to find the estimate. But here it is not $0$. Don't know how to proceed.

If you've seen the Cauchy integral formula for derivatives of a complex analytic function $f(z)$ then you can use that to get a bound on both the first and second derivatives of $f(z)$ at $z=0$. The Cauchy integral formula for derivatives (on the unit disk for simplicity) says that if $f(z)$ is analytic on $\bar{D}$ and $|f(z)|\leq M$ then 
$$f^{(n)}(0)=\frac{n!}{2\pi i}\int_{|z|=1}\frac{f(z)}{z^{n+1}}\,dz$$
Now if you parametrize the boundary by $z=e^{i\theta}$ you can get
$$f^{(n)}(0)=\frac{n!}{2\pi i}\int_0^{2\pi}f\left(e^{i\theta}\right)e^{-in\theta}i\,d\theta$$
So that
$$|f^{(n)}(0)|\leq\frac{n!}{2\pi}\int_0^{2\pi}|f\left(e^{i\theta}\right)|\,d\theta$$
and therefore
$$|f^{(n)}(0)|\leq n!M$$
since $|f\left(e^{i\theta}\right)|\leq M$ by hypothesis.
Now apply that for $n=2$ on your second derivative of $e^f$ (so define $f$ above as $f:=e^f$) and you'll have an upper bound with $M=\left|e^f\right|\leq e^{|f|}=e$ and hence $\left|\left(e^f\right)''\right|(0)\leq 2e<6$  
