Finding general solution $\frac{d^2f}{dx^2}$ for an implicit function Let's say I have a function $F:\mathbb R \times \mathbb R \rightarrow \mathbb R$, which meets the requirements of the implicit function theorem. Now I would like to find a general expression for:
$$\frac{d^2f}{dx^2}$$
From the theorem I know that $$\frac{df}{dx} = - \Big(\frac{\partial F}{\partial y} \Big )^{-1} \frac{\partial F}{\partial x}$$
So I went on to say that $$\frac{d^2f}{dx^2} = -  \frac{\frac{d}{dx}(\frac{\partial F}{\partial x})*\frac{\partial F}{\partial y}-\frac{\partial F}{\partial x}*\frac{d}{dx}(\frac{\partial F}{\partial y})}{(\frac{\partial F}{\partial y})^2}$$
By using the quotient rule. Is this correct, and if so: How can I simplify this any further? I found online that $$\frac{d^2f}{dx^2}=-\frac
{\frac{\partial^2 F}{\partial x^2}\left(\frac{\partial F}{\partial y}\right)^2
-2·\frac{\partial^2 F}{\partial x\partial y}·\frac{\partial F}{\partial y}·\frac{\partial F}{\partial x}
+\frac{\partial^2 F}{\partial y^2}\left(\frac{\partial F}{\partial x}\right)^2}
{\left(\frac{\partial F}{\partial y}\right)^3}$$
which is correct I guess, but I do not quite know how to get there. Any help is greatly appreciated!

HINT:
Let $\displaystyle F_x(x,y)=\frac{\partial F(x,y)}{\partial x}$ and $\displaystyle F_y(x,y)=\frac{\partial F(x,y)}{\partial y}$.
$$\begin{align}
\frac{d}{dx}F_x(x,f(x))&=F_{xx}(x,f(x))+F_{xy}(x,f(x))\frac{df(x)}{dx}\\\\
&=F_{xx}(x,f(x))-F_{xy}(x,f(x))\frac{F_x(x,f(x))}{F_y(x,f(x))}
\end{align}$$
