Answer
$z_x=-\dfrac{x}{3z}$ and $z_y=-\dfrac{2y}{3z}$
Work Step by Step
We are given that $x^2+2y^2+3z^2=1$
Re-arrange as: $x^2+2y^2+3z^2-1=0$
Consider $F(x,y,z)=x^2+2y^2+3z^2-1=0$
$F_x=2x$ and $F_y=4y$ and $F_z=6z$
Use Equation 7 which is: $z_x=-\dfrac{F_x}{F_z}$ and $z_y=-\dfrac{F_y}{F_z}$
$z_x=-\dfrac{F_x}{F_z}=-\dfrac{2x}{6z}$ and $z_y=-\dfrac{F_y}{F_z}=-\dfrac{4y}{6z}$
Hence, we have $z_x=-\dfrac{x}{3z}$ and $z_y=-\dfrac{2y}{3z}$
