Let $F:\mathbb{R}^2 \rightarrow \mathbb{R}$ be three times continuously differentiable in some open neighborhood $\mathcal{U}$ of $(0,0)$. Suppose that $F(0,0) = F\_x(0,0) = F\_y(0,0) = F\_{xy}(0,0) = 0$ and that $F\_{yy} \not = 0$ and $F\_{xx}(0,0)/F\_{yy}(0,0) < 0$.


Obviously I cannot apply the implicit function theorem. Under which circumstances can I still express y as a function of x locally aoround (0,0)? It seems as though it should be $y'(0) = \pm \sqrt{-\frac{F\_{xx}(0,0)}{F\_{yy}(0,0)}}$...


EDIT: Fixed typo.



Implicit function theorem at a singular point?
This is a job for the Morse lemma. The second degree Taylor polynomial of $F(x,y)$ has the form $ax^2+by^2$ where $ab<0$. (You said $>0$ but that can't be what you meant.) The Morse Lemma says, for a sufficiently smooth function of several variables, that if it has zero constant and linear parts and a nondegenerate quadratic part then there is a change of variables making the function purely quadratic. In the case at hand (two variables and indefinite quadratic part) that means that $F=uv$ where $u(x,y)$ and $v(x,y)$ have no constant term and have linearly independent linear parts whose product is $ax^2+by^2$. Now the implicit function theorem applies to $u$ and to $v$ and you find that the solution set of $F(x,y)=0$ is locally the union of the graphs of two smooth functions, one with positive slope and one with negative slope.


EDIT Here is a proof of the Morse Lemma: 


First, in one variable it says that if $f(0)=f\_x(0)=0$ and $f\_{xx}(0)\neq 0$ then in a neighborhood of $x=0$ $f(x)$ is plus or minus the square of a smooth function. This is true because the vanishing of $f(0)$ means that $f(x)=xg(x)$ for some smooth $g$, and the vanishing of $f\_x(0)=g(0)$ means that $g(x)=xh(x)$ for some smooth $h$, and the nonvanishing of $f\_{xx}(0)=2g\_x(0)=2h(0)$ means that $h$ is locally plus or minus a square.


Now suppose that $F(x,y)$ is such that $F$, $F\_x$, and $F\_y$ vanish at $x=y=0$ but $F\_{yy}$ does not, and suppose that the homogeneous quadratic part of the Taylor series is nondegenerate. The implicit function theorem applied to $F\_y$ shows that $F\_y=0$ along the graph of a smooth function $y=k(x)$ with $k(0)=0$. Define $f(x)=F(x,k(x))$. Then $F(x,y)-f(x)$ is such that both it and its $y$-derivative vanish along that graph. Therefore we may write $F(x,y)-f(x)=(y-k(x))^2H(x,y)$ for some smooth $H$. Furthermore $2H(0,0)=F\_{yy}(0,0)\neq 0$, so $H$ is plus or minus the square of a smooth function. Now $F(x,y)$ is the sum of $F(x,y)-f(x)$, which is the square of a smooth function, and $f(x)$, which must also be plus or minus the square of a smooth function by the one-variable case. 


And so on.


