I tried to use Liouville theorem to show that function is a constant function first, but I found it useless. Then I found that modulus of 1/$f$(z) is less than a bounded function, but here also lead to a problem that I cannot make sure the 1/$f$(z) is holomorphic except origin point. I wonder how to solve this problem.
$zf(z)\to 0$ as $z\to 0$ so the singularity at $z=0$ is removable. So $f$ is entire.
Now, using the Cauchy estimate, for $z_0\in \mathbb C$,
$f'(z_0)=\frac1{2\pi i}\int_{|z-z_0|=R}\frac{f(z)}{(z-z_0)^2}\,\mathrm{d}z\Rightarrow |f'(z_0)|\le \frac{|f(z_0+Re^{it})|}{R^{2}}\le \frac{1}{R^{3/2}}+\frac{1}{R^{5/2}}\to 0$ as $R\to \infty$
so $f'(z_0)=0$ and $f$ is constant.
$f$ has a removable singularity at $0$ because $zf(z) \to 0$ as $z \to 0$. Let $g(z)=\frac {f(z)-f(0)} z$ for $z \neq 0$ and $f'(0)$ for $z=0$. The $g$ is entire and the hypothesis tells you that is it bounded. Hence $g$ is a constant and $f(z)=az+b$ for some $a,b$. Rest should be clear.
