Answer
$z_x=-\dfrac{x}{3z}$ and $z_y=-\dfrac{2y}{3z}$
Work Step by Step
Recall Equation 7 such as: $z_x=-\dfrac{F_x}{F_z}$ and $z_y=-\dfrac{F_y}{F_z}$ ...(1)
Given: $x^2+2y^2+3z^2=1$
$\implies x^2+2y^2+3z^2-1=0$
Let us consider that $F(x,y,z)=x^2+2y^2+3z^2-1=0$
$F_x=2x\\ F_y=4y\\F_z=6z$
Now, we have $z_x=-\dfrac{F_x}{F_z}=-\dfrac{2x}{6z}=-\dfrac{x}{3z}$ and $z_y=-\dfrac{F_y}{F_z}=-\dfrac{4y}{6z}=-\dfrac{2y}{3z}$
and $z_x=-\dfrac{x}{3z}\\z_y=-\dfrac{2y}{3z}$
