If $f$ is entire and $\lim_\limits{z\to\infty} \frac{f(z)}{z} = 0$ show that $f$ is constant

There is a flaw in the proof. Note that for the given $\epsilon>0$, $|f(z)|<\epsilon |z|$ for $|z|>n_0$.

Therefore, on $|\zeta|=R$, $|f(\zeta)|<\epsilon |\zeta|=\epsilon R$. Then, we can write for $R>z$

$$\begin{align} |f'(z)|&=\left|\frac{1}{2\pi i}\oint_{|\zeta|=R}\frac{f(\zeta)}{(\zeta-z)^2}\,d\zeta\right|\\\\ &\le \frac{1}{2\pi}\frac{\epsilon R}{(R-|z|)^2}\,(2\pi R)\\\\ &=\epsilon \frac{R^2}{(R-|z|)^2} \end{align}$$

As $R\to \infty$, we find that for any $\epsilon>0$, there exists a number $n_0$, such that whenever $|z|>n_0$, $|f'(z)|<\epsilon$. We can conclude from this only that

$$\lim_{z\to \infty}f'(z)=0$$

Another approach is to write $f(z)$ in terms of its Taylor series. Then, we see that

$$\begin{align} \lim_{z\to \infty}\frac{f(z)}{z}&=\lim_{z\to \infty}\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}z^{n-1}\\\\ &=\lim_{z\to \infty}\left(\frac{f(0)}{z}+f'(0)+\frac12 f''(0)z+\cdots \right)\\\\ &=0 \end{align}$$

only if all terms in the series are zero, except possibly $f(0)$. Therefore, $f(z)$ must be a constant.

A simpler solution with the use of maximum modulus theorem.

Define $g : \mathbb{C} \rightarrow \mathbb{C}$, $g(z)=\begin{cases} \frac{f(z)-f(0)}{z-0}, & \text{if }z \neq 0 \\ f'(0), & \text{if }z=0 \end{cases}$

So $g$ is also an entire function by addition and quotient of entire functions. $\lim_{z\to\infty}|g(z)|=\lim_{z\to\infty} |\frac{f(z)}{z}|=0$. So $\exists L$ such as $\forall z > L, |g(z)|< \epsilon$. In particular it is true on the circle of center $0$ and of radius $L$.

With the maximum modulus theorem we have that the $max$ of $g$ on the closed discus of center $0$ and of radius $L$ is equal to the $max$ of $g$ on the circle of center $0$ and of radius $L$. So $\forall z \in \mathbb{C}, |g(z)|< \epsilon$. So $\forall z \in \mathbb{C}, g(z)=0$. So $\forall z \in \mathbb{C},f(z)=f(0)$. So $f$ is constant.

As @Dr.Mv said, we have that $\lim_{z\rightarrow\infty}f'(z)=0$, which means that $f'$ is bounded. By Liouville's theorem $f'$ is constant and, since $\lim_{z\rightarrow\infty}f'(z)=0$, $f'=0$. Thus , $f$ is constant.