# Question concerning implicit function theorem

I am studying the implicit function theorem. I have a question about the condition $$F_y(x_0,y_0)\neq 0.$$

More precisely, let $$w=F(x,y)$$ be a $$C^1$$ function on an open rectangle $$R=(a,b) \times (c,d)$$. Consider $$F(x,y)=C,$$ where $$C \in \mathbb{R}$$ is fixed, and $$F(x_0,y_0)=C$$ for some $$(x_0,y_0) \in R.$$

By implicit function theorem, if $$F_y(x_0,y_0)\neq 0,$$ $$\exists$$ a $$C^1$$ function $$y=y(x)$$ near $$(x_0,y_0)$$ satisfying $$F(x,y(x))=C.$$

If $$\exists$$ a differential function $$y=y(x)$$ near $$(x_0,y_0)$$ satisfying $$F(x,y(x))=C,$$ then, by chain rule, $$F_x(x_0,y_0)+F_y(x_0,y_0)y'(x_0)=0,$$ which implies that the case $$F_x(x_0,y_0)\neq 0$$ and $$F_y(x_0,y_0)=0$$ is impossible.

My question: Is it possible that $$F_x(x_0,y_0)= F_y(x_0,y_0)=0$$?

I would be grateful if you give any comment for my question. Thanks in advance.

Sure. Let's focus on the case $$(x_0,y_0)=(0,0)$$ (up to translation of the plane this is general). There can be several related phenomena producing $$F_x=F_y=0$$:

1) There could be several branches of $$F(x,y)=0$$ near $$(0,0)$$. For example $$F(x,y)=x^2-y^2$$ has $$F_x=F_y=0$$ but of course there is $$y(x)=x$$ such that $$F(x,y(x))=0$$. In fact for $$y(x)=-x$$ we also have $$F(x,y(x))=0$$, which is the second branch (similarly for $$F(x,y)=xy$$, with the second branch $$x=0$$ vertical). Note that the extra branches may be not visible over real numbers, for example $$F(x,y)=(x-y)(x^2+y^2)$$ which over reals is equivalent to $$y=x$$, but over complex numbers has other solutions $$y=ix$$ and $$y=-ix$$.

2) The multiplicity of the single branch could be more than 1. For example $$F(x,y)=y^2$$ with $$y(x)=0$$ or $$F(x,y)=(x-y)^2$$ with $$y(x)=x$$ (or $$F(x,y)=y^5$$ etc.). This is like multiple branches, but with two or more "different" branches becoming equal.