$f$ is entire, satisfying $$|f(z)|\le \frac1{|\text{Re}(z)|}.$$ Show that $f$ is identically zero.
It suffices to show boundedness. Since $f$ is bounded on the real line, then if $f$ is unbounded on the complex plane, the infinity must be an essential singularity, which means, for an arbitrarily chosen point $z_0\in\Bbb C$, we can pick a sequence $z_n$ whose moduluses tend to infinity with $f(z_n)\to z_0$, and it follows that $|z_0|\le \frac1{|\text{Re}(z_n)|}$. This is the best I can do.
I'm not sure if I'm going in the correct direction. It however seems unlikely for me to deal with the problem on the $y$-strip in this way.