= Solution
Solve the two algebraic <constraints> for the squared coordinates:
$$
x^2=\frac{z^2+u^2}{3},\qquad y^2=\frac{z^2-2u^2}{3}.
$$
On any nonsingular real branch, <implicit differentiation> at fixed $u$ gives $(\partial x/\partial z)_u=z/(3x)$ and $(\partial y/\partial z)_u=z/(3y)$. Differentiating $z^2=2x^2+y^2$ with the other original coordinate fixed gives $(\partial z/\partial x)_y=2x/z$ and $(\partial z/\partial y)_x=y/z$. Therefore
$$
\boxed{\left(\frac{\partial x}{\partial z}\right)_u\left(\frac{\partial z}{\partial x}\right)_y+\left(\frac{\partial y}{\partial z}\right)_u\left(\frac{\partial z}{\partial y}\right)_x=\frac z{3x}\frac{2x}z+\frac z{3y}\frac yz=\frac23+\frac13=1.}
$$
This is the <multivariable chain rule> applied to $z(x(z,u),y(z,u))=z$. These formulas assume the relevant denominators are nonzero and the chosen variables are valid local coordinates.
Back to article page
