1. We are given that for all $x,y\in \mathbb{R}$, $|f(x)-f(y)|\le (x-y)^2$ and $f$ is differentiable with $f(0)=0$.

We need to find $f(1)$.

2. Since $f$ is differentiable, let us use the given inequality to study the derivative.

Take any fixed $x\in \mathbb{R}$ and let $y=x+h$ where $h\neq 0$. Then $|f(x+h)-f(x)|\le h^2.$ Divide both sides by $|h|$: $\left|\frac{f(x+h)-f(x)}{h}\right|\le |h|.$

3. Now let $h\to 0$. Since $f$ is differentiable at $x$, $\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=f'(x).$ From the inequality above, $\left|\frac{f(x+h)-f(x)}{h}\right|\le |h| \to 0.$ Hence by squeeze theorem, $|f'(x)|=0 \implies f'(x)=0$ for every $x\in \mathbb{R}$.

4. Therefore, $f'(x)=0$ for all $x$, so $f$ must be a constant function on $\mathbb{R}$.

Let $f(x)=c$. Given $f(0)=0$, we get $c=0.$ So, $f(x)=0 \quad \text{for all } x.$ In particular, $f(1)=0.$

5. Checking options:

• A: $-1$ ❌
• B: $0$ ✅
• C: $2$ ❌
• D: $1$ ❌

Therefore the correct answer is B.