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Plugging in $x=1$ in the second equality gives $f(1)=1$. By induction, it follows that $f(x)=x$ for all positive integers $x$.

Note that $f(x+k)=f(x)+k$ for positive integer $k$. So, $f(m+\frac{k}{m})^2 = (f(\frac{k}{m})+m)^2$. By the second F.E condition, $f(m+\frac{k}{m})^2 = f(m^2+2k+\frac{k^2}{m^2})=m^2+2k+f(\frac{k^2}{m^2})$. Equating the two gives us $f(\frac{k}{m})=\frac{k}{m}$.