Answer
(a) ${F_x} = {z^2} + y$, ${\ \ }$ ${F_y} = 2yz + x$, ${\ \ }$ ${F_z} = 2xz + {y^2}$
(b) $\frac{{\partial z}}{{\partial x}} = - \frac{{{z^2} + y}}{{2xz + {y^2}}}$, ${\ \ \ }$ $\frac{{\partial z}}{{\partial y}} = - \frac{{2yz + x}}{{2xz + {y^2}}}$
Work Step by Step
(a) We have $F\left( {x,y,z} \right) = x{z^2} + {y^2}z + xy - 1$.
The partial derivatives are
${F_x} = {z^2} + y$, ${\ \ }$ ${F_y} = 2yz + x$, ${\ \ }$ ${F_z} = 2xz + {y^2}$
(b) Using Eq. (7) we obtain
$\frac{{\partial z}}{{\partial x}} = - \frac{{{F_x}}}{{{F_z}}}$, ${\ \ \ }$ $\frac{{\partial z}}{{\partial y}} = - \frac{{{F_y}}}{{{F_z}}}$
$\frac{{\partial z}}{{\partial x}} = - \frac{{{z^2} + y}}{{2xz + {y^2}}}$, ${\ \ \ }$ $\frac{{\partial z}}{{\partial y}} = - \frac{{2yz + x}}{{2xz + {y^2}}}$
