The point $(1,-1,2)$ lies on each of the surfaces $$x^2(y^2+z^2)=5,\ (x-z)^2+y^2=2.$$ Prove that in a neighborhood of that point, the intersection of the surfaces can be described as a smooth curve $z=f(x),\ y=g(x)$ and determine the direction of the tangent to it at $(1,-1,2)$.
From substituting $y^2=2-(x-z)^2$ into the first equation I got $$x^4-2x^3z-2x^2+5=0,$$ so I believe the intersection is given by that equation. Now I need to apply the Implicit function theorem somehow, I suppose. It's natural to consider the map $F: \mathbb R^3\to \mathbb R$ given by $$(x,y,z)\mapsto x^4-2x^3z-2x^2+5.$$ Am I supposed to apply the IFT twice? Note that $F_z\ne 0$ at the specified point, so I can write $z=\rho(x,y)$ with $F(x,y,\rho(x,y))=0$ in a neighborhood, but $F_y=0$. So this doesn't give much. How should I proceed?
Hint:
First solve both equations wrt $z$ as a function of $x,y$, i.e find $z_i (x,y)$ for both equations, $i=1,2$.
Then equate $z_1 = z_2$, and get a new function $F(x,y) = $z_1 (x,y) - z_2 (x,y) = 0.$$
Now, solve this equation wrt $x$,or $y$.