Answer
$\frac{{\partial z}}{{\partial x}} = - \frac{{{{\rm{e}}^z} - 1}}{{x{{\rm{e}}^z} + {{\rm{e}}^y}}}$
Work Step by Step
We are given $x{{\rm{e}}^z} + z{{\rm{e}}^y} = x + y$.
Write $F\left( {x,y,z} \right) = x{{\rm{e}}^z} + z{{\rm{e}}^y} - x - y = 0$. So, $z$ is defined implicitly by $F\left( {x,y,z} \right) = 0$.
The partial derivatives are
${F_x} = {{\rm{e}}^z} - 1$, ${\ \ \ }$ ${F_y} = z{{\rm{e}}^y} - 1$, ${\ \ \ }$ ${F_z} = x{{\rm{e}}^z} + {{\rm{e}}^y}$
Using Eq. (7) of Section 15.6, we get
$\frac{{\partial z}}{{\partial x}} = - \frac{{{F_x}}}{{{F_z}}} = - \frac{{{{\rm{e}}^z} - 1}}{{x{{\rm{e}}^z} + {{\rm{e}}^y}}}$
