Answer
$$\frac{{\partial z}}{{\partial x}} = - \frac{{2x + y}}{{y + 2z}}{\text{ and }}\frac{{\partial z}}{{\partial x}} = - \frac{{x + 2y + z}}{{y + 2z}}$$
Work Step by Step
$$\eqalign{
& {x^2} + xy + {y^2} + yz + {z^2} = 0 \cr
& {\text{Letting }}F\left( {x,y,z} \right) = {x^2} + xy + {y^2} + yz + {z^2} \cr
& {\text{Calculate }}{F_x}\left( {x,y,z} \right),{\text{ }}{F_y}\left( {x,y,z} \right){\text{ and }}{F_z}\left( {x,y,z} \right) \cr
& {F_x}\left( {x,y,z} \right) = \frac{\partial }{{\partial x}}\left[ {{x^2} + xy + {y^2} + yz + {z^2}} \right] = 2x + y \cr
& {F_y}\left( {x,y,z} \right) = \frac{\partial }{{\partial y}}\left[ {{x^2} + xy + {y^2} + yz + {z^2}} \right] = x + 2y + z \cr
& {F_z}\left( {x,y,z} \right) = \frac{\partial }{{\partial z}}\left[ {{x^2} + xy + {y^2} + yz + {z^2}} \right] = y + 2z \cr
& {\text{Using Theorem 13}}.{\text{8}},{\text{ you have}} \cr
& \frac{{\partial z}}{{\partial x}} = - \frac{{{F_x}\left( {x,y,z} \right)}}{{{F_z}\left( {x,y,z} \right)}} = - \frac{{2x + y}}{{y + 2z}} \cr
& \frac{{\partial z}}{{\partial y}} = - \frac{{{F_y}\left( {x,y,z} \right)}}{{{F_z}\left( {x,y,z} \right)}} = - \frac{{x + 2y + z}}{{y + 2z}} \cr
& \cr
& \frac{{\partial z}}{{\partial x}} = - \frac{{2x + y}}{{y + 2z}}{\text{ and }}\frac{{\partial z}}{{\partial x}} = - \frac{{x + 2y + z}}{{y + 2z}} \cr} $$
