Is there a proof for $\frac{\mathrm{d}y}{\mathrm{d}x} = - \frac{\frac{\partial }{\partial x}}{\frac{\partial }{\partial y}}$?

Question

I've read that $$\frac{\mathrm{d}y}{\mathrm{d}x} = - {\frac{\frac{\partial}{∂x}}{\frac{∂}{∂y}}}$$

For example, if $f(x,y) = 2xy + y^2,$

$$\frac{\partial f(x,y)}{\partial x} = 2y \ \ \text{ and }\ \ \frac{\partial f(x,y)}{\partial y} = 2x+2y, $$

so $\displaystyle \frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{2y}{2x+2y}$

Is there a proof for this assertion? I can't seem to find one.

Answer 1

The comment by GEdgar points out the essential point already. First you are given a function $f(x, y)$ of two variables $x$, $y$. At this point $x, y$ are independent variables, so when you write $$\frac{\mathrm d y}{\mathrm dx},$$ you are thinking of $y=y(x)$ as a function of $x$. That is, you are restricting to a curve in the $x-y$ plane, but which curve?

Indeed if you are restricting to a level curve, then one can prove your assertion: by definition, if $(x, y(x))$ represent a portion of a level curve, then $$f(x, y(x)) = c$$ for some fixed constant $c$. Differentiating the above with respect to $x$ and using Chain rule, we have

$$ \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{\mathrm dy}{\mathrm dx} = 0, $$ which implies

$$ \frac{\mathrm dy}{\mathrm dx} =- \frac{\frac{\partial f}{\partial x}}{\frac{\partial f}{\partial y}}.$$