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A function $f: A \to B$ is injective if $f(x_1, y_1) = f(x_2, y_2)$ implies $(x_1, y_1) = (x_2, y_2)$ for all $(x_1, y_1), (x_2, y_2) \in Z \times Z$.
Let $(x_1, y_1)$ and $(x_2, y_2)$ be two elements in $Z \times Z$ such that $f(x_1, y_1) = f(x_2, y_2)$.
Given $f(x, y) = (2y, 3x)$, the equality becomes: $$(2y_1, 3x_1) = (2y_2, 3x_2)$$
By comparing the corresponding components: $$2y_1 = 2y_2 \implies y_1 = y_2$$ $$3x_1 = 3x_2 \implies x_1 = x_2$$
Since $f(x_1, y_1) = f(x_2, y_2)$ leads to $(x_1, y_1) = (x_2, y_2)$, the function is injective.
Final Answer: The function is injective.
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