If $f^2$ is Riemann Integrable is $f$ always Riemann Integrable?
Yes, there exists such functions. Think of $$ f(x) = \begin{cases} 1 & \text{ if } x \in \mathbb Q \\ -1 & \text{ if } x \notin \mathbb Q. \end{cases} $$ It is well-known that $f$ is not Riemann-integrable over any interval $[a,b]$ (just compute the Riemann lower/upper sums). But $f^2 = 1$ is very integrable. =)
$$f=2\cdot\mathbf 1_{[a,b]\cap\mathbb Q}-1$$