All points at which the surfaces $x^2+y^2+z^2-1=0$ and $x^2+y^2-z^2-2y=0$ are intersect orthogonally $f: x^2+y^2+z^2-1=0$
$g: x^2+y^2-z^2-2y=0$
I set these two surfaces equal to each other to solve for the intersection, getting 
$y=(1-z^2)/2$...then attempted to insert this value of $y$ in terms of $z$ back into the original equations, took the gradient of each, and set the dot product of the two gradients equal to $0$. My result was that $4x^2+16z^2(z-1)(z-2)=0$ which doesn't seem to have a great solution that exists on my two surfaces. Further, I realize that the intersection I found at the beginning doesn't make much sense as it is a parabola in the $yz$ plane and f is the unit sphere. I also tried equating both surfaces to where they equal $z^2$ and got the intersection 
$x^2+(y-1/2)^2=5/8$...but am unsure of how to proceed in terms of orthogonality. Where did I go wrong? thanks in advance.

If we set $f(x,y,z)=x^2+y^2+z^2-1=0$ and $g(x,y,z)=x^2+y^2-z^2-2y=0$ then $$\nabla f=(2x,2y,2z),~~\nabla g=(2x,2y-2,-2z)$$ If we want to find that points so they are lying on the a curve in which $(\nabla f\cdot\nabla g) (x,y,z)=0$. This gives us $$4(x^2+y^2-z^2)-4y=0$$ But $x^2+y^2-z^2=2y$ on the intersection so $$4(x^2+y^2-z^2)-4y=0\to y=0$$ Now we should think of the possible solution(s) of the following system:
$$ \left\{
        \begin{array}{ll}
            x^2+z^2=1 \\
            x^2-z^2=0
        \end{array}
    \right.$$
