Solutions

Notice that

$$\begin{align*} |f(x)-f(y)|\le |x-y|^n &\implies \frac{|f(x)-f(y)|}{|x-y|}\le |x-y|^{n-1} \\ &\implies \left|\frac{f(x)-f(y)}{x-y}\right|\le |x-y|^{n-1} \\ &\implies -|x-y|^{n-1}\le \frac{f(x)-f(y)}{x-y}\le |x-y|^{n-1}. \end{align*}$$

By the squeeze theorem, since for any $y$ we have

$$\lim_{x\to y}-|x-y|^{n-1}=\lim_{x\to y}|x-y|^{n-1}=0,$$

then

$$f'(y)=\lim_{x\to y}\frac{f(x)-f(y)}{x-y}=0.$$

Therefore, $f$ is constant.