TITLE: How to show that $\frac{dy}{dx}=-\frac{F_x(x,y)}{F_y(x,y)}$
QUESTION [7 upvotes]: Suppose that $y$ is defined implicitly as a function $y(x)$ by an equation on the form $F(x,y)=0$. I'm trying to show that
$$\frac{dy}{dx}=-\frac{F_x(x,y)}{F_y(x,y)},$$
but I don't know where to start. Can someone please give me a hint? 
Both $y(x)$ and $F(x,y)$ are differentiable and $F_y(x,y)\neq 0$.

REPLY [10 votes]: The key point is to use the chain rule. From $F(x,y(x))=0$ get $F_x(x,y)\cdot1+F_y(x,y)y'(x)=0$.
The derivation above has a geometric interpretation: The gradient of $F$ is orthogonal to the level curve $F=0$. Hence, it's orthogonal to the tangent vector.

REPLY [9 votes]: What happens when you take a small example, say you want to find $\mathrm{d}y/\mathrm{d}x$ of $x^2+y^2=9$, you have:
$$\begin{align}
2x+\frac{\mathrm{d}y}{\mathrm{d}x}2y&=0\\
\frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{x}{y}
\end{align}$$
Now you have $F(x,y)=0$, differentiation gives:
$$\begin{align}
F_x(x,y)+F_y(x,y)\frac{\mathrm{d}y}{\mathrm{d}x}&=0\\
\frac{\mathrm{d}y}{\mathrm{d}x}&=-\frac{F_x(x,y)}{F_y(x,y)}
\end{align}$$