Solve $(x^2+y^2+2z^2)^{1/2}=\cos z$ for $z$ near $(0,1,0)$ 
Can the equation $(x^2+y^2+2z^2)^{1/2}=\cos z$ be solved uniquely for $z$ in terms of $x$ and $y$ near $(0,1,0)$?

Put $F(x,y,z)=(x^2+y^2+2z^2)^{1/2}-\cos z$. Then $F(0,1,0)=0$,
$$
\nabla F(x,y,z)=(x(x^2+y^2+2z^2)^{-1/2},y(x^2+y^2+2z^2)^{-1/2},2z(x^2+y^2+2z^2)^{-1/2}+\sin z)
$$
and $\nabla F(0,1,0)=(0,1,0)$. Hence by the implicit function theorem, $y$ is a function of $x$ and $z$ near $(0,1,0)$.
But we cannot conclude the solvability of $z$ from this. Since for all $z$ sufficiently close to $0$, $F(x,y,z)=0$ if and only if $F(x,y,-z)=0$, $z$ cannot be solved as a function of $x$ and $y$. Am I correct? How to make this rigorous?
 A: For this kind of problems, you need to use the multidimensional Taylor series  taking directional derivatives.
Truncated to the second order, this gives
$$\frac{1}{2}x^2+y+\frac{3 }{2}z^2-1=0 \tag 1$$ Checking at $x=\frac 1{10}$, $y=\frac 9{10}$, we have
$$\sqrt{2 z^2+\frac{41}{50}}-\cos (z)=0$$ which, expanded, gives
$$0=\frac{1}{10} \left(\sqrt{82}-10\right)+\left(\frac{1}{2}+5
   \sqrt{\frac{2}{41}}\right) z^2+O\left(z^4\right)$$ that is to say
$$z^2=\frac{41 \left(10-\sqrt{82}\right)}{5 \left(41+10 \sqrt{82}\right)}=0.0589$$
while $(1)$ gives
$$z^2=\frac{19}{300}=0.0633$$
Expanded to the next order
$$\left(-\frac{x^2 y}{2}+x^2+y-1\right)+\left(\frac{5}{2}-y\right) z^2=0$$ which, for the used values would give
$$z^2=\frac{189}{3200}=0.0591$$ the exact solution being $z^2=0.0604$
A: In order to prove that $F(x,y,z)=0$ cannot be solved uniquely for $z$ in terms of $x$ and $y$ near $(0,1,0)$, we show that for any $\frac{1}{\sqrt{3}}<y<1$ the equation
$$f(z):=F(0,y,z)=(y^2+2z^2)^{1/2}-\cos(z)=0$$
has at least two solutions $\pm z$ with $z\in(0,y)$.
In fact $f(0)=|y|-\cos(0)=y-1<0$ and $$f( y)=(y^2+2y^2)^{1/2}-\cos(y)=\sqrt{3}|y|-\cos(y)\geq \sqrt{3}y-1>0.$$
The existence of $z$ follows by the Intermediate value theorem applied to the even continuous $f$ and the interval $(0,y)$. Note that $z\to 0^+$ as $y\to 1^-$.
