The question asks you to solve this using the jacobian method. From the question, we get that: $X = VU$ and $Y = U(V-1)$ and so the jacobian yields: $|J| = |U|$.
So we get: $f_{UV}(u,v) =|J| \cdot f_{XY}(uv,u(v-1))$. Since X and Y are i.i.d, $f_{XY}(x,y) = 1/4$, if x $\in (0,2)$ and y $\in (0,2)$.
So the final result is: $f_{UV}(u,v) =|u|/4$ if $0<uv<2$ and $0<u(v-1)<2$.
But both regions $0<uv<2$ and $0<u(v-1)<2$ described by this joint distribution have infinite area, so is the final joint distribution valid? Or maybe the jacobian method cannot be applied in this situation?
First of all there is a mistake. $Y = U(1-V)$ and not $U(V-1)$.
Now, $f_{UV}(u,v) = u \cdot f_{XY}(uv,u(1-v))$. I removed the absolute value as $u$ cannot be negative. You will see why as you go through the answer.
As you said, $0 \lt uv \lt 2$ and $0 \lt u (1-v) \lt 2$
From $0 \lt uv \lt 2$, observe that $u$ and $v$ must have the same sign.
From $0 \lt u(1-v) \lt 2$, $v$ cannot be greater thant $1$ as $u$ and $v$ canno]t have different signs.
Also $u$ and $v$ cannot be negative as for $u, v \lt 0$, $u(1-v) \lt 0$
Now for $0 \lt u \lt 2$ and $0 \lt v \lt 1$, there are no restrictions and any values of $u, v$ work.
Also note intersection of $uv = u (1-v) \implies v = \frac{1}{2}, u = 4$
So for $2 \lt u \lt 4$, we have
$\frac{u-2}{u} \lt v \lt \frac{2}{u}$ and $v$ is always less than $1$
For $u \gt 4$, we get contradiction in value of $v$ from both equations.
So support of the distribution is given by,
$$0 \lt v \lt 1, 0 \lt u \lt 2$$ $$\frac{u-2}{u} \lt v \lt \frac{2}{u}, 2 \lt u \lt 4$$