Let $\widehat{f}(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi}f(x)e^{-inx}dx.$ Then, since $f$ is infinitely differentiable, we have $\widehat{f^{(k)}}(n)=\left(in\right)^{k}\widehat{f}(n).$ Parsevals theorem tells us that $\sum_{n=-\infty}^{\infty}\left|\widehat{f}(n)\right|^{2}=\frac{1}{2\pi}\int_{-\pi}^{\pi}\left|f(x)\right|^{2}dx,$ and hence combining the fact that $|f^{(k)}(x)|\leq 1$ along with the formula for $\widehat{f^{(k)}}$ in terms of $\hat{f}(k)$ we have $\sum_{n=-\infty}^{\infty}n^{2k}\left|\widehat{f}(n)\right|^{2}\leq1.$ Taking $k$ to infinity implies the only non zero terms are $n=-1,0,1,$ and we have that $f(x)=\widehat{f}(1)e^{ix}+\hat{f}(0)+\hat{f}(-1)e^{-ix}.$ From here we just have to play around with the other givens until it works out. Since f^'(0)=1, and $f(x)\leq 1$, we can argue that $\hat{f}(0)=0$. This then allows us to solve for $f$.