Answer
$z_x=-\dfrac{x}{3z} \\ z_y=-\dfrac{2y}{3z}$
Work Step by Step
Re-write as $F(x,y,z)=x^2+2y^2+3z^2-1=0$
Now, $F_x=2x ; F_y=4y; F_z=6z$
Since, $z_x=-\dfrac{F_x}{F_z}$ and $z_y=-\dfrac{F_y}{F_z}$
So, $$z_x=-\dfrac{F_x}{F_z}=-\dfrac{2x}{6z} \\ z_y=-\dfrac{F_y}{F_z}=-\dfrac{4y}{6z}$$
