= Solution
Differentiate both algebraic <constraints> at fixed $y$ or fixed $x$:
$$
z_x=\frac{2x}z,\quad z_y=\frac yz,\qquad u_x=\frac xu,\quad u_y=-\frac yu.
$$
The <multivariable chain rule> for $\phi(x,y)=\psi(z,u)$ therefore gives
$$
\left(\frac{\partial\phi}{\partial x}\right)_y=\frac{2x}z\left(\frac{\partial\psi}{\partial z}\right)_u+\frac xu\left(\frac{\partial\psi}{\partial u}\right)_z,\qquad\left(\frac{\partial\phi}{\partial y}\right)_x=\frac yz\left(\frac{\partial\psi}{\partial z}\right)_u-\frac yu\left(\frac{\partial\psi}{\partial u}\right)_z.
$$
Divide the first identity by $x$ and the second by $y$, and add. The $\psi_u$ terms cancel, leaving
$$
\boxed{\frac1x\left(\frac{\partial\phi}{\partial x}\right)_y+\frac1y\left(\frac{\partial\phi}{\partial y}\right)_x=\frac3z\left(\frac{\partial\psi}{\partial z}\right)_u.}
$$
The <Jacobian determinant> of $(x,y)\mapsto(z,u)$ is $-3xy/(zu)$, so the calculation is valid on real branches with $xyzu\ne0$. This explicitly supplies the local independence assumption underlying the change of coordinates.
Back to article page
