1. Given conditions

We are given a twice differentiable function $f:\mathbb{R}\to\mathbb{R}$ such that $f(0)=0,\quad f(1)=0,\quad f'(0)=0.$ We must determine which statement is necessarily true.

2. Use Rolle's theorem on $[0,1]$

Since $f$ is twice differentiable, it is continuous on $[0,1]$ and differentiable on $(0,1)$. Also, $f(0)=f(1)=0.$ So by Rolle's theorem, there exists some $c\in(0,1)$ such that $f'(c)=0.$

But we are also given $f'(0)=0.$ Thus, $f'(0)=f'(c)=0.$

3. Apply Rolle's theorem to $f'$ on $[0,c]$

Because $f$ is twice differentiable, $f'$ is continuous on $[0,c]$ and differentiable on $(0,c)$. Since $f'(0)=f'(c)=0,$ Rolle's theorem applied to $f'$ gives some $\xi\in(0,c)\subset(0,1)$ such that $f''(\xi)=0.$

Hence there exists at least one point in $(0,1)$ where $f''$ vanishes.

So option B is true.

4. Check the other options

Option A: $f''(x)\ne 0$ for every $x\in(0,1)$

This contradicts what we just proved: there exists $\xi\in(0,1)$ with $f''(\xi)=0.$ So A is false.

Option C: $f''(0)=0$

This need not be true. Consider $f(x)=x^2(x-1)=x^3-x^2.$ Then $f(0)=0,\quad f(1)=0,\quad f'(x)=3x^2-2x,$ so $f'(0)=0.$ But $f''(x)=6x-2 \implies f''(0)=-2\ne 0.$ So C is false.

Option D: $f''(x)=0$ at every point $x\in(0,1)$

If this were true, then $f$ would be linear on $(0,1)$. With $f'(0)=0$, that would force a constant behavior, which is not necessary. The same counterexample works: $f(x)=x^2(x-1),\quad f''(x)=6x-2,$ which is not identically zero on $(0,1)$. So D is false.

5. Conclusion

The only statement that must hold for all such twice differentiable functions is: $\boxed{\text{B: } f''(x)=0 \text{ for some } x\in(0,1).}$