We define $(a, b) \diamond(c, d)=(a c-b d, a d+b c)$.

Determine all pairs $(x, y)$ for which $(x, 3) \diamond(x, y)=(6,0)$.
From the definition, $(x, 3) \diamond(x, y)=\left(x^{2}-3 y, x y+3 x\right)$.

Since $(x, 3) \diamond(x, y)=(6,0)$, then $x^{2}-3 y=6$ and $x y+3 x=0$.

From the second equation $x(y+3)=0$ so $x=0$ or $y=-3$.

If $x=0$, then $x^{2}-3 y=6$ gives $0-3 y=6$ or $y=-2$.

If $y=-3$, then $x^{2}-3 y=6$ gives $x^{2}+9=6$ or $x^{2}=-3$, which is not possible.

Therefore, $(x, y)=(0,-2)$.

ANSWER: $(0,-2)$