Answer
$-\frac{1}{2}.$
Work Step by Step
We have $$
\operatorname{Jac}(G)=\frac{\partial(x, y)}{\partial(u, v)}=\left|\begin{array}{ll}
{\frac{\partial x}{\partial u}} & {\frac{\partial x}{\partial v}} \\
{\frac{\partial y}{\partial u}} & {\frac{\partial y}{\partial v}}
\end{array}\right|=\left|\begin{array}{ll}
{ v/u} & {\ln u} \\
{2u/v} & {-u^2/v^2}
\end{array}\right| =-(u/v)-(2u(\ln u)/v).
$$
Now at the point $(u,v)= (1,2)$, we have $
\operatorname{Jac}(G)=-(1/2)-0=-\frac{1}{2}.$
