To prove that if the last digit of the number $x^2 + xy + y^2$ is zero, then the last two digits of this number are zeros, we will proceed as follows:

1. Understanding the Problem:
- We are given that the last digit of $x^2 + xy + y^2$ is zero. This implies that $x^2 + xy + y^2 \equiv 0 \pmod{10}$.
- We need to show that $x^2 + xy + y^2 \equiv 0 \pmod{100}$.

2. Prime Factorization of 10:
- Note that $10 = 2 \times 5$. Therefore, $x^2 + xy + y^2 \equiv 0 \pmod{10}$ implies $x^2 + xy + y^2 \equiv 0 \pmod{2}$ and $x^2 + xy + y^2 \equiv 0 \pmod{5}$.

3. Divisibility by 2:
- Since $x^2 + xy + y^2 \equiv 0 \pmod{2}$, we know that $x$ and $y$ must both be even or both be odd. This is because the sum of squares and a product of integers will be even if and only if all terms are even or all terms are odd.

4. Divisibility by 5:
- Since $x^2 + xy + y^2 \equiv 0 \pmod{5}$, we need to consider the possible values of $x$ and $y$ modulo 5. We will use the fact that the quadratic residues modulo 5 are $0, 1, 4$.

5. Combining the Conditions:
- We need to show that $x$ and $y$ must be such that $x^2 + xy + y^2 \equiv 0 \pmod{25}$ as well. This will ensure that $x^2 + xy + y^2 \equiv 0 \pmod{100}$.

6. Detailed Calculation:
- Consider $x \equiv 0 \pmod{5}$ and $y \equiv 0 \pmod{5}$. Then $x = 5a$ and $y = 5b$ for some integers $a$ and $b$.
- Substituting these into the expression, we get:
$$x^2 + xy + y^2 = (5a)^2 + (5a)(5b) + (5b)^2 = 25a^2 + 25ab + 25b^2 = 25(a^2 + ab + b^2)$$
- Since $25(a^2 + ab + b^2)$ is clearly divisible by 25, it follows that $x^2 + xy + y^2 \equiv 0 \pmod{25}$.

7. Conclusion:
- Since $x^2 + xy + y^2 \equiv 0 \pmod{4}$ and $x^2 + xy + y^2 \equiv 0 \pmod{25}$, by the Chinese Remainder Theorem, we have $x^2 + xy + y^2 \equiv 0 \pmod{100}$.

Therefore, the last two digits of $x^2 + xy + y^2$ are zeros.