= Solution
{c}
In a <partial derivative>, the variable held fixed matters. Holding $u$ fixed gives $x_v=-v$. To find $v_y$, hold $x$ fixed and invert the <Jacobian matrix>:
$$
\begin{pmatrix}dx\\dy\end{pmatrix}=\begin{pmatrix}u&-v\\v&u\end{pmatrix}\begin{pmatrix}du\\dv\end{pmatrix},\qquad \begin{pmatrix}du\\dv\end{pmatrix}=\frac1{u^2+v^2}\begin{pmatrix}u&v\\-v&u\end{pmatrix}\begin{pmatrix}dx\\dy\end{pmatrix}.
$$
Thus
$$
\boxed{\left.\frac{\partial x}{\partial v}\right|_u=-v,\qquad \left.\frac{\partial v}{\partial y}\right|_x=\frac{u}{u^2+v^2}.}
$$
The second formula holds on each local inverse branch away from $(u,v)=(0,0)$, where the <Jacobian determinant> vanishes. These are <planar parabolic coordinates>; the pairs $(u,v)$ and $(-u,-v)$ represent the same $(x,y)$, so no global choice of inverse is implicit.
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