The correct answer is $\phi(u, v) = 1$.
To see this, let us consider a double integral over the region $R$ in the $xy$-plane:
$$\iint_R f(x, y) \, dxdy$$
We can write this integral in terms of the new variables $u$ and $v$ as follows:
$$\iint_R f(uv, v/u) \phi(u, v) \, dudv$$
where $\phi(u, v)$ is a Jacobian of the transformation from the $xy$-plane to the $uv$-plane.
The Jacobian is a determinant that tells us how much the area of a small rectangle in the $xy$-plane changes when we transform it to the $uv$-plane. In this case, the Jacobian is equal to $1$, so we can write the integral as follows:
$$\iint_R f(uv, v/u) \, dudv = \iint_R f(x, y) \, dxdy$$
Therefore, $\phi(u, v) = 1$.