$f$ is bounded because $f$ is integrable and satisfies your inequality. Notice that the integral is monotonically increasing in $x$. In other words:
For some finite $C$.
As a hint for the exercise, try integrating both sides of the inequality from $0$ to $x$ and notice that if $f$ is bounded, then $\int_0^y\int_0^xf(t)dtdx\leq yC$. On the other hand, $|f(y)|\leq\int_0^y|f(x)|dx$. What happens if you do three integrals...four integrals...