Answer
1. At the point $\left( {3,2,1} \right)$
$\frac{{\partial z}}{{\partial x}}{|_{\left( {x,y,z} \right) = \left( {3,2,1} \right)}} = - \frac{3}{{11}}$
$\frac{{\partial z}}{{\partial y}}{|_{\left( {x,y,z} \right) = \left( {3,2,1} \right)}} = \frac{1}{{22}}$
2. At the point $\left( {3,2, - 1} \right)$
$\frac{{\partial z}}{{\partial x}}{|_{\left( {x,y,z} \right) = \left( {3,2, - 1} \right)}} = \frac{3}{{11}}$
$\frac{{\partial z}}{{\partial y}}{|_{\left( {x,y,z} \right) = \left( {3,2, - 1} \right)}} = - \frac{1}{{22}}$
Work Step by Step
Write $F\left( {x,y,z} \right) = {z^4} + {z^2}{x^2} - y - 8$.
The partial derivatives are
${F_x} = 2{z^2}x$, ${\ \ }$ ${F_y} = - 1$, ${\ \ }$ ${F_z} = 4{z^3} + 2z{x^2}$
Using Eq. (7) we obtain
$\frac{{\partial z}}{{\partial x}} = - \frac{{{F_x}}}{{{F_z}}}$, ${\ \ \ }$ $\frac{{\partial z}}{{\partial y}} = - \frac{{{F_y}}}{{{F_z}}}$
$\frac{{\partial z}}{{\partial x}} = - \frac{{2{z^2}x}}{{4{z^3} + 2z{x^2}}}$, ${\ \ \ }$ $\frac{{\partial z}}{{\partial y}} = \frac{1}{{4{z^3} + 2z{x^2}}}$
1. At the point $\left( {3,2,1} \right)$
$\frac{{\partial z}}{{\partial x}}{|_{\left( {x,y,z} \right) = \left( {3,2,1} \right)}} = - \frac{{2\cdot{1^2}\cdot3}}{{4\cdot{1^3} + 2\cdot1\cdot{3^2}}} = - \frac{3}{{11}}$
$\frac{{\partial z}}{{\partial y}}{|_{\left( {x,y,z} \right) = \left( {3,2,1} \right)}} = \frac{1}{{4\cdot{1^3} + 2\cdot1\cdot{3^2}}} = \frac{1}{{22}}$
2. At the point $\left( {3,2, - 1} \right)$
$\frac{{\partial z}}{{\partial x}}{|_{\left( {x,y,z} \right) = \left( {3,2, - 1} \right)}} = - \frac{{2\cdot{{\left( { - 1} \right)}^2}\cdot3}}{{4\cdot{{\left( { - 1} \right)}^3} + 2\cdot\left( { - 1} \right)\cdot{3^2}}} = \frac{3}{{11}}$
$\frac{{\partial z}}{{\partial y}}{|_{\left( {x,y,z} \right) = \left( {3,2, - 1} \right)}} = \frac{1}{{4\cdot{{\left( { - 1} \right)}^3} + 2\cdot\left( { - 1} \right)\cdot{3^2}}} = - \frac{1}{{22}}$
