= Solution
For this <change of variables>,
$$
u_x=2x,\quad u_y=-2y,\quad v_x=2y,\quad v_y=2x,
$$
and $u_{xx}+u_{yy}=0=v_{xx}+v_{yy}$. The <second derivative chain rule> therefore gives
$$
\begin{aligned}
f_{xx}+f_{yy}
&=g_{uu}(4x^2+4y^2)+2g_{uv}(4xy-4xy)+g_{vv}(4y^2+4x^2)\\
&=4(x^2+y^2)(g_{uu}+g_{vv}).
\end{aligned}
$$
Thus the transformation rule for the <Laplacian> is
$$
\boxed{\Delta_{x,y}f=4(x^2+y^2)\Delta_{u,v}g.}
$$
For $f=x+y$, its <Laplacian> is zero. At every point with $(x,y)\ne(0,0)$, division by $4(x^2+y^2)$ proves
$$
\boxed{g_{uu}+g_{vv}=0.}
$$
This conclusion applies to each local inverse branch of the <change of variables>: its <Jacobian determinant> is $4(x^2+y^2)$, so such a branch exists away from the origin by the <inverse function theorem>. A single globally defined $g$ cannot represent $x+y$ on the entire plane, since $(x,y)$ and $(-x,-y)$ give the same $(u,v)$ but generally opposite values of $x+y$. Nor can a differentiable $g$ represent this $f$ in a neighborhood of the origin: the <multivariable chain rule> there would give $f_x=f_y=0$, contradicting $f_x=f_y=1$. The requested <harmonic function> conclusion is therefore a local statement on nonsingular branches.<Calculus, probability, linear algebra, discrete mathematics, logic and proof, and models of computation examination paper by the University of Oxford (2003)>[Printed p. 2, Question 2(a)(ii)]{ref}
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