As we know, for implicit differentiation we have $\frac{\partial y}{\partial x}=-\frac{F_x}{F_y}$
but if I use a trick that $\frac{\partial y}{\partial x}=\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}=\frac{\partial F}{\partial x}\frac{\partial y}{\partial F}=\frac{\partial y}{\partial x}$ as $\partial F$ cancels each other out, so it is like $\frac{\partial y}{\partial x}=\frac{F_x}{F_y}$ without the "minus", where did I go wrong?
This is a classic case of "differentials are not fractions" and is a marked downside of Leibniz notation. While the notation $\frac{d y}{d x }$ or $\frac{\partial F}{\partial y}$ looks like a fraction, we can't simply "cancel the $\partial F$" out. In effect, you have just proved a difference between differentials and fractions.
For another, consider the chain rule in several variables. Let $F= F(x,y)$ and let $x=x(s,t)$ and $y=y(s,t)$. Then $$\frac{\partial F}{\partial t} = \frac{\partial F}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial F}{\partial y} \frac{\partial y}{\partial t}$$
If we were allowed to cancel terms like you suggest, then we would have that $\frac{\partial F}{\partial t} = 2 \frac{\partial F}{\partial t}$, which is clearly not true.