= Solution
At fixed $z$, differentiating $2x^2+y^2=z^2$ gives $(\partial x/\partial y)_z=-y/(2x)$. Also, $x^2=(z^2+u^2)/3$ gives $(\partial x/\partial u)_z=u/(3x)$. At fixed $y$, eliminate $x^2$ to write $u^2=(z^2-3y^2)/2$, giving $(\partial u/\partial z)_y=z/(2u)$. Finally, the previous part gave $(\partial y/\partial z)_u=z/(3y)$. Thus
$$
\boxed{\left(\frac{\partial x}{\partial y}\right)_z\left(\frac{\partial y}{\partial z}\right)_u+\left(\frac{\partial x}{\partial u}\right)_z\left(\frac{\partial u}{\partial z}\right)_y=-\frac y{2x}\frac z{3y}+\frac u{3x}\frac z{2u}=0.}
$$
The held-fixed subscripts matter: substituting a reciprocal <partial derivative> while silently changing what is held fixed would not preserve these cancellations.
Back to article page
