(a) Every perfect cube $k^3$ of a positive integer is super special because we can write
$$k^3 = k^3 \frac{x^3 + 2y^3}{x^3 + 2y^3} = \frac{(kx)^3 + 2(ky)^3}{x^3 + 2y^3}$$
for some positive integers $x, y$.

(b) Observe that $2014 = 2 \cdot 19 \cdot 53$. If $2014$ is super special, then we have,
$$x^3 + 2y^3 = 2014(u^3 + 2v^3) \qquad (1)$$
for some positive integers $x, y, u, v$. We may assume that $x^3 + 2y^3$ is minimal with this property. Now, we will use the fact that if $19$ divides $x^3 + 2y^3$, then it divides both $x$ and $y$. Indeed, if $19$ does not divide $x$, then it does not divide $y$ too. The relation $x^3 \equiv -2y^3 \pmod{19}$ implies $(x^3)^6 \equiv (-2y^3)^6 \pmod{19}$. The latter congruence is equivalent to $x^{18} \equiv 2^6y^{18} \pmod{19}$. Now, according to Fermat's Little Theorem, we obtain $1 \equiv 2^6 \pmod{19}$, that is $19$ divides $63$, not possible.
It follows $x = 19x_1, y = 19y_1$, for some positive integers $x_1$ and $y_1$. Replacing in (1) we get
$$19^2(x_1^3 + 2y_1^3) = 2 \cdot 53(u^3 + 2v^3) \qquad (2)$$
i.e. $19|u^3 + 2v^3$. It follows $u = 19u_1$ and $v = 19v_1$, and replacing in (2) we get
$$x_1^3 + 2y_1^3 = 2014(u_1^3 + 2v_1^3).$$
Clearly, $x_1^3 + 2y_1^3 < x^3 + 2y^3$, contradicting the minimality of $x^3 + 2y^3$.