The pair of lanthanides in which both elements have high third-ionization energy is:

We need to find the pair of lanthanides where both elements have high third-ionization energy (IE$$_3$$).

The third ionization energy of a lanthanide involves removing the third electron. A high IE$$_3$$ occurs when removing the third electron disrupts an especially stable electron configuration (half-filled or fully-filled 4f subshell).

Europium (Eu): Electronic configuration is [Xe] 4f$$^7$$ 6s$$^2$$. After losing two electrons (from 6s): Eu$$^{2+}$$ = [Xe] 4f$$^7$$. The third ionization requires removing an electron from the half-filled 4f$$^7$$ shell, which is very stable. Hence, Eu has a high IE$$_3$$.

Ytterbium (Yb): Electronic configuration is [Xe] 4f$$^{14}$$ 6s$$^2$$. After losing two electrons (from 6s): Yb$$^{2+}$$ = [Xe] 4f$$^{14}$$. The third ionization requires removing an electron from the completely filled 4f$$^{14}$$ shell, which is very stable. Hence, Yb has a high IE$$_3$$.

Checking the other options:

Option A (Dy, Gd): Gd has configuration [Xe] 4f$$^7$$ 5d$$^1$$ 6s$$^2$$, so Gd$$^{2+}$$ = [Xe] 4f$$^7$$ 5d$$^1$$. The third electron comes from 5d, not disrupting the 4f$$^7$$, so IE$$_3$$ is not exceptionally high. Dy$$^{2+}$$ = [Xe] 4f$$^{10}$$, which has no special stability.

Option B (Lu, Yb): Lu has configuration [Xe] 4f$$^{14}$$ 5d$$^1$$ 6s$$^2$$, so Lu$$^{2+}$$ = [Xe] 4f$$^{14}$$ 5d$$^1$$. The third electron comes from 5d, so IE$$_3$$ is not especially high.

Option D (Eu, Gd): Eu has high IE$$_3$$, but Gd does not (as explained above).

The correct answer is Option C: Eu, Yb.