Answer
$$\frac{{\partial z}}{{\partial x}} = - \frac{{{z^2}}}{{{x^2}}}$$
Work Step by Step
$$\eqalign{
& {\text{Let }}\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1 \cr
& {x^{ - 1}} + {y^{ - 1}} + {z^{ - 1}} = 1 \cr
& {\text{Find }}\frac{{\partial z}}{{\partial x}},{\text{ differentiate both sides}} \cr
& \frac{\partial }{{\partial x}}\left[ {{x^{ - 1}}} \right] + \frac{\partial }{{\partial x}}\left[ {{y^{ - 1}}} \right] + \frac{\partial }{{\partial x}}\left[ {{z^{ - 1}}} \right] = \frac{\partial }{{\partial x}}\left[ 1 \right] \cr
& - {x^{ - 2}} - {z^{ - 2}}\frac{{\partial z}}{{\partial x}} = 0 \cr
& {\text{Solve for }}\frac{{\partial z}}{{\partial x}} \cr
& - {z^{ - 2}}\frac{{\partial z}}{{\partial x}} = {x^{ - 2}} \cr
& \frac{{\partial z}}{{\partial x}} = - \frac{{{x^{ - 2}}}}{{{z^{ - 2}}}} \cr
& \frac{{\partial z}}{{\partial x}} = - \frac{{{z^2}}}{{{x^2}}} \cr} $$
