I just read that:
If $z=f(x,y)=c$, be the equation of a curve, then the slope of the tangent to the curve at any point (x,y), is given by $$m=\frac {dy}{dx}=-\frac{\frac{\partial z}{\partial x}}{\frac {\partial z}{\partial y}}$$
I don't see how the minus sign creeps in here.(Of course I don't have a proof, but the - sign is against intuition).
A proof(or a link to a simple proof) would be nice, and an intuitive explanation would be nicer. Thanks for help.
As $z=f(x,y)=c$ $$\bigtriangledown z=\frac{\partial z}{\partial x}dx+\frac {\partial z}{\partial y}dy=0$$ so$$ \frac {dy}{dx}=-\frac{\frac{\partial z}{\partial x}}{\frac {\partial z}{\partial y}} $$
This can be intuitively understood by realizing that we are calculating the Slope of the Tangents to the Level Curves of a Surface.
Level Curves (and thus their Tangents) always run Perpendicular to the Gradient Vector of the Surface.
The Gradient Vector has Slope ${f_y\over f_x}$.
The Slope of a Line Perpendicular to another has as Slope the Negative Reciprocal of the other Line's Slope.
Therefore the Implicit Derivative is $-{f_x\over f_y}$ because this is the Negative Reciprocal of the Slope of the Surface $f(x,y)$'s Gradient.
From the chain rule, we get $$ 0 = \begin{pmatrix} \frac{\partial{f}}{\partial{x}} & \frac{\partial{f}}{\partial{y}}\end{pmatrix} \begin{pmatrix} 1 \\ \frac{dy}{dx}\end{pmatrix}, $$ which yields a linear equation $$ \frac{\partial{f}}{\partial{y}} \frac{dy}{dx} = - \frac{\partial{f}}{\partial{x}}. $$
Now, we apply Cramer's rule to find $\frac{dy}{dx}.$ That is, $$ \frac{dy}{dx} = \frac{|- \frac{\partial{f}}{\partial{x}}|}{|\frac{\partial{f}}{\partial{y}}|} = \frac{-| \frac{\partial{f}}{\partial{x}}|}{|\frac{\partial{f}}{\partial{y}}|} = - \frac{ \frac{\partial{f}}{\partial{x}}}{\frac{\partial{f}}{\partial{y}}}, $$ where $|A|$ denotes the determinant of $A.$ Therefore, the minus sign is rather natural from the joint perspective of multivariable calculus and linear algebra.
