Problem: Can the equation $(x^2+y^2+z^2)^{1/2}=\cos z$ can be solved uniquely for $z$ in terms of $x,y$ in the neighborhood of the point $(0,1,0)$?
Let $F(x,y,z)=(x^2+y^2+z^2)^{1/2}-\cos z,$ then $F(0,1,0)=0$ and $F'_{z}(0,1,0)=0$. Thus, the Implicit function theorem does not work.
I want to show that $F$ cannot be expressed as a graph $z=f(x,y)$ near $(0,1,0)$. Pick $x=0$, then $(y^2+z^2)^{1/2}=\cos z$. The graph shows that for all $y$ in the neighborhood of $1$, there exists two different corresponding values of $z$, but I don't know how to prove it.
If $(x,y,z)$ is a solution, then so is $(x,y,-z)$ so it will suffice to show that there $is$ a solution for $z\neq 0$ in any neighborhood of $(0,1,0).$ Then, we have to show that $\sqrt{x^2+y^2+z^2}-\cos z=0$ when $x,y,z$ are included in an arbitrary neighborhood $N$ of $(0,1,0).$
Now, for all $0<\epsilon<1$, this expression is negative if $x=z=0,$ and $y=1-\epsilon.$ On the other hand, if $y=1,\ x=0$ and $z= \epsilon,$ the expression is positive. We may choose $\epsilon$ small enough so that $P=(0,1-\epsilon,0)$ and $Q=(0,1,\epsilon)$ are both contained in $N$. And because $N$ is convex, so is the line segment $\overline {PQ}.$
Consider $f(x,y,z)=\sqrt{x^2+y^2+z^2}-\cos z.$ It is continuous on the connected set $S=\{(1-t)P+tQ:\ 0\le t\le 1\}$ so $f(S)$ is connected. We have just shown that $f(P)<0$ and $f(Q)>0$ so connectedness of $f(S)$ implies that there is a $0<t_0<1$ such that $f((1-t_0)P+t_0Q)=0.$ The result now follows because $z\neq 0$ on the interior of $S$.