Infinitely differentiable implicit function

Question

I have an infinitely differentiable function $F(x,y) : [0,1]^2 \rightarrow \mathbb{R}$. I know that for every $x \in [0,1]$ there is a unique $y$ for which $F(x,y)=0$. Can I claim that the function $y(x)$ given by $F(x,y) = 0$ is infinitely differentiable and differentiate $F(x,y) = 0$ w.r.t $x$?

Answer 1

No. Take $F(x,y)=x^2+y^2$ for $0 \le x \le 1$ and $0 \le y \le 1$. Since $y \ge 0$ we get

$y(x)=\sqrt{1-x^2}$.

$y$ is not differentiable at $x=1$.