= Solution
On a branch with fixed signs of $u,v$ and $uv\ne0$, use $u^2=x+y$ and $v^2=y-x$. Differentiating these identities, or inverting the <Jacobian matrix>, gives
$$
\boxed{u_x=\frac1{2u},\quad u_y=\frac1{2u},\quad
v_x=-\frac1{2v},\quad v_y=\frac1{2v}.}
$$
The inverse <Jacobian matrix> is therefore
$$
\frac{\partial(u,v)}{\partial(x,y)}
=\begin{pmatrix}1/(2u)&1/(2u)\\-1/(2v)&1/(2v)\end{pmatrix}.
$$
These are local <partial derivatives>. Choosing the signs specifies an <inverse function> on $y>|x|$; the original map has four preimages there and has no single global inverse. At $u=0$ or $v=0$, the displayed inverse derivatives are not finite.
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