Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 2 / 9 / a / ii / 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 $(\partial y/\partial z{)}_u=z/(3y)$. ThusThe held-fixed subscripts matter: substituting a reciprocal partial derivative while silently changing what is held fixed would not preserve these cancellations.