Let $F(x,y)$ be a $C^2$-function defined on a neighborhood of $(x_0,y_0)\in \Bbb R^2$. Suppose that $F(x_0,y_0)=0, F_y(x_0,y_0)\ne0$. By Implicit Function Theorem, the equation $F(x,y)=0$ defines a $C^2$-curve $C:y=f(x)$ on a neighborhood of $(x_0,y_0)$. My questions are as follows:
(1) Show that the tangent line to $C$ at $(x_0,y_0)$ is parallel to $x$-axis if and only if $F_x(x_0,y_0)=0$.
(2) Under the assumption of (1), show that if $F_y(x_0,y_0)F_{xx}(x_0,y_0)<0$ then there exists a neighborhood $U$ of $(x_0,y_0)$ such that $C \cap U \subset \{(x_0,y_0)\in \Bbb R^2\mid y \ge y_0\}$.
Firstly, I know the following formulae: $y'=f'(x)=-\frac{F_x}{F_y}$ and $y''=-\frac{F_{xx}F_y^2-2F_{xy}F_{x}F_{y}+F_{yy}F_x^2}{F_y^3}$. For (1), the tangent line to $C$ at $(x_0,y_0)$ is parallel to $x$-axis $\iff y'(x_0,y_0)=0 \iff F_x(x_0,y_0)=0$. I guess the formula of $y''$ will play some role in the proof of (2) but I can't figure it out.
Any help would be much appreciated.
Here's a hint to get you started:
In the formula for $y''(x_0)$, you have $F_x=0$, which leaves only the factors $F_{xx}$ and $F_y$. This should be enough for you to determine the sign of $y''(x_0)$, and because of continuity it will keep that sign in a neighbourhood of $x_0$ too.