To solve this problem, we evaluate each Roman numeral by applying the definition of the operation $\(a \oplus b=a^2-a b\)$ under the constraint that $\(x \neq 0\)$ and $\(y \neq 0\)$.

Evaluating Statement I: $\(x \oplus y\)$
According to the definition:

$$
\(x \oplus y=x^2-x y\)
$$


To see if this can be zero, we set the expression to zero:

$$
\(\begin{aligned}
& x^2-x y=0 \\
& x(x-y)=0
\end{aligned}\)
$$


Since the problem states $\(x \neq 0\)$, we can divide both sides by $\(x\)$ :

$$
\(x-y=0 \Longrightarrow x=y\)
$$


Since $x$ and $y$ can be any non-zero numbers, they can certainly be equal (e.g., $\(x=2, y=2\)$ ).
Therefore, Statement I can be zero.

Evaluating Statement II: $\(x y \oplus y\)$
Substitute $\(a=x y\)$ and $\(b=y\)$ into the definition:

$$
\(\begin{gathered}
x y \oplus y=(x y)^2-(x y)(y) \\
=x^2 y^2-x y^2
\end{gathered}\)
$$


Set the expression to zero:

$$
\(x^2 y^2-x y^2=0\)
$$

Factor out the common term $\(x y^2\)$ :

$$
\(x y^2(x-1)=0\)
$$


Since we are told $x \neq 0$ and $\(y \neq 0\)$, the term $\(x y^2\)$ cannot be zero. Thus, we must have:

$$
\(x-1=0 \Longrightarrow x=1\)
$$


Since $x$ can be 1 , Statement II can be zero (e.g., when $\(x=1\)$ and $\(y=5\)$ ).

Evaluating Statement III: $\(x \oplus(x+y)\)$
Substitute $\(a=x\)$ and $\(b=(x+y)\)$ into the definition:

$$
\(x \oplus(x+y)=x^2-x(x+y)\)
$$


Distribute the $\(-x\)$ :

$$
\(\begin{gathered}
=x^2-\left(x^2+x y\right) \\
=x^2-x^2-x y \\
=-x y
\end{gathered}\)
$$


For this to be zero, either $\(x\)$ or $\(y\)$ must be zero. However, the prompt explicitly states $\(x y \neq 0\)$. Therefore, Statement III cannot be zero.

Conclusion
Only statements I and II can be equal to zero.
Correct Answer: B. I and II only