Answer
$\frac{d^2y}{dx^2}=\frac{F_{xx}F^2_{y}-2F_{xy}F_xF_y+F_{yy}F^2_{x}}{F^3_y}$
Work Step by Step
Before we start, we first derive some things we need
$\frac{\partial}{\partial x}F_x=\frac{\partial^2F}{\partial x^2}+\frac{\partial^2F}{\partial y\partial x}\frac{dy}{dx}=F_{xx}-F_{yx}\frac{F_x}{F_y}$
$\frac{\partial}{\partial x}F_y=\frac{\partial^2F}{\partial x\partial y }\frac{dx}{dx}+\frac{\partial^2F}{\partial y^2}\frac{dy}{dx}=F_{xy}-F_{yy}\frac{F_x}{F_y}$
Therefore
$\frac{d^2y}{dx^2}=-\frac{\frac{\partial F_x}{\partial x
}F_y-\frac{\partial F_y}{\partial x}Fx}{(F_y)^2}=-\frac{F_{xx}F_{y}-F_{yx}F_x-F_{xy}F_x+\frac{F_{yy}F^2_{x}}{F_y}}{F^2_y}$
Since $F$ has continuous second derivative, by Clairaut's Theorem, $F_{yx}=F_{xx}$, thus, we have
$\frac{d^2y}{dx^2}=-\frac{F_{xx}F^2_{y}-2F_{xy}F_xF_y+F_{yy}F^2_{x}}{F^3_y}$
