I have a question about the implicit function theorem. I have one equation with 3 variables and I have to know if we can solve the equation uniquely for $y$. But I don't know when that is allowed and how I can do it.
It is about this equation: $$\sqrt{x^2+y^2+z^2} - \cos(z) = 0$$ in a neighborhood of $(0,1,0)$.
When you have a single equation, $$ f(x,y,z) = 0,$$ the implicit function theorem is fairly straightforward to apply. (Here, we're assuming that $f$ is continuously differentiable.)
Suppose that the point $(a,b,c)$ obeys $f(a,b,c) = 0 $, and suppose we want to prove that there exists a continuously differentiable function $g(x,z)$, defined on a neighbourhood of $(a,c)$, such that $g(a,c) = b$, and such that $$f(x,g(x,z),z) = 0.$$
By the implicit function theorem, all we have to do is show that $$ \frac {\partial f}{\partial y}(a,b,c) \neq 0.$$
So in your case, we apply this with $f(x,y,z) = \sqrt{x^2 + y^2 + z^2} - \cos z$ and $(a,b,c) = (0,1,0)$.
Anyway, in this example, it's not hard to solve for $g(x,z)$ directly: I believe the answer is $$ g(x,z) = + \sqrt{\cos^2 z - x^2 - z^2},$$ valid on an open neighbourhood of $(0,1,0)$ chosen to be small enough such that the expression inside the square root is positive throughout this neighbourhood.