For all twice differentiable functions $$f : \mathbb{R} \to \mathbb{R}$$, with $$f(0) = f(1) = f'(0) = 0$$,

We are given a twice differentiable function $$f : \mathbb{R} \to \mathbb{R}$$ with $$f(0) = 0$$, $$f(1) = 0$$, and $$f'(0) = 0$$.

Step 1: Apply Rolle's Theorem to $$f$$ on $$[0, 1]$$.

Since $$f$$ is differentiable (hence continuous) on $$[0, 1]$$ and $$f(0) = f(1) = 0$$, by Rolle's Theorem there exists $$c_1 \in (0, 1)$$ such that:

$$f'(c_1) = 0$$

Step 2: Apply Rolle's Theorem to $$f'$$ on $$[0, c_1]$$.

Since $$f$$ is twice differentiable, $$f'$$ is differentiable on $$(0, c_1)$$ and continuous on $$[0, c_1]$$. We have:

$$f'(0) = 0 \quad \text{and} \quad f'(c_1) = 0$$

By Rolle's Theorem applied to $$f'$$, there exists $$c_2 \in (0, c_1) \subset (0, 1)$$ such that:

$$f''(c_2) = 0$$

Step 3: Verify with a counterexample that other options fail.

Consider $$f(x) = x^2(x - 1) = x^3 - x^2$$. One verifies:

$$f(0) = 0, \quad f(1) = 0, \quad f'(x) = 3x^2 - 2x \implies f'(0) = 0$$

The second derivative is $$f''(x) = 6x - 2$$, so:

$$f''(0) = -2 \neq 0 \quad (\text{Option C is false})$$

$$f''(x) = 0 \text{ only at } x = \tfrac{1}{3} \quad (\text{Option A is false, Option D is false})$$

Conclusion:

For every twice differentiable function satisfying the given conditions, $$f''(x) = 0$$ for some $$x \in (0, 1)$$. This is guaranteed by the double application of Rolle's Theorem.

The correct answer is Option B: $$f''(x) = 0$$, for some $$x \in (0, 1)$$.