If $f$ is a real-valued differentiable function satisfying $|f(x)-f(y)|\le (x-y{)}^2$ for $x,y\in \mathbb{R}$ and $f(0)=0$, then $f(1)$ equals:

The given inequality is $|f(x)-f(y)|\le (x-y{)}^2$.Divide by $|x-y|$ (assuming $x\ne y$):
$$\left|\frac{f(x)-f(y)}{x-y}\right|\le |x-y|$$
Taking the limit as $y\to x$:
$$|f^{\prime }(x)|\le \underset{y\to x}{\lim }|x-y|$$
$$|f^{\prime }(x)|\le 0⟹f^{\prime }(x)=0$$
Since the derivative is zero for all $x$, $f(x)$ must be a constant function.$f(x)=C$.Given $f(0)=0$, then $C=0$.Thus, $f(x)=0$ for all $x$, and $f(1)=0$.