Misuse of inverse function theorem

Question

Let $U=\{\binom{u}{v}\in \mathbb{R}^2|0<v<u\}$. Let $f:U\to\mathbb{R}^2, f(u,v)=(u+v,uv)$. Show that $f$ has a global reverse function, find $g=f^{-1}$ and its domain.

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Not a valid solution: For any $(u,v)\in\mathbb{R}^2$, $$ \\ Df(u,v)=\begin{pmatrix} 1& 1\\ v& u \end{pmatrix} \ne0 \Leftrightarrow u\ne v \ $$ then $Df(u,v)$ is reversable for any $(u,v)\in U$. Thus, $\forall (u,v)\in U$, $$ \\ g(u,v)=\frac{\operatorname{adj}(Df(u,v))}{\operatorname{det}(Df(u,v))}=\frac{1}{u-v}\cdot \begin{pmatrix} u & -1 \\ -v & 1 \end{pmatrix} \ $$ I understand this is not the solution, but I don't understand where is the mistake.

Answer 1

You've found the inverse of $Df$, not the inverse of $f$.