Understanding partial derivatives of multi-variable functions I'm having trouble convincing myself of the validity of the following set of steps in differentiation of a multi-variable function under a constraint:
Suppose $f(x,y) = 0$
The above constraint implicitly makes $x$ dependent on $y$.
I can do the following operation 
\begin{align}
\frac{df(x,y)}{dx} = 0
\end{align}
Since the derivative is not partial, $y$ is allowed to change with changing $x$ and the constraint $f(x,y) = 0$ can still be satisfied. I know that taking partial derivative wouldn't be legal since that would make $y$ constant and the constraint will no longer be satisfied. 
How do I prove this formally? That
\begin{align}
\frac{{\partial f(x,y)}}{{\partial x}}
\end{align}
is invalid given that $f(x,y) = 0$.
 A: Well, that's two different concepts: the partial derivative and the total derivative.
Assuming $x$ and $y$ are the independent variables in the function $f(x,y)$, we can define its partial derivatives
$$\frac{\partial f}{\partial x}=\frac{\partial f(x,y)}{\partial x} \quad \text{and} \quad \frac{\partial f}{\partial y}=\frac{\partial f(x,y)}{\partial y}.$$
Assuming $y$ is also defined implicitly as a function of $x$, we can say that we have two functions: a function of two variables $f(x,y)$, where we can still treat $y$ is one of the independent variables; and a function of a single variable $x$ defined as $g(x)=f(x,y(x))$. I think what you're trying to find is the derivative of this new function:
$$\frac{dg}{dx}=\frac{df}{dx}=\frac{df(x,y(x))}{dx}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\cdot\frac{dy}{dx}.$$
Note that in this context, both notations $\displaystyle \frac{\partial f}{\partial x}$ and $\displaystyle \frac{df}{dx}$ make sense but mean different things.
Moreover, when an equation such as $f(x,y)=0$ defines $y$ as an implicit function of $x$, this is the basis for the implicit differentiation method for finding the derivative of $y$ with respect to $x$:
$$f(x,y)=0 \implies \frac{df}{dx}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\cdot\frac{dy}{dx}=0 \implies \frac{dy}{dx}=-\frac{\partial f/\partial x}{\partial f/\partial y}.$$
A: I've always thought it from the definition of the derivative:
$$ \frac{df}{dx} = \lim_{\Delta x\rightarrow 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$
Multivariable versions are just slightly different:
$$ \frac{\partial f}{\partial x} = \lim_{\Delta x \rightarrow 0}\frac{f(x+\Delta x,y)-f(x,y)}{\Delta x} $$
and
$$ \frac{\partial f}{\partial y} = \lim_{\Delta y \rightarrow 0}\frac{f(x,y + \Delta y)-f(x,y)}{\Delta y} $$
Obviously (x,y) is a parameter in all the versions...
Important property is obviously the linearity:
$$ L(a+b) = L(a)+L(b) $$ $$ L(c a) = c L(a)$$, where you can choose $L=\frac{\partial f}{\partial x}$ or $L=\frac{\partial f}{\partial y}$
