Partial derivative identity help Please can some one give me the proof for this, ( I think y and x can be written as parametric equations in terms of C)(∂y/∂x) = (∂y/∂c) / (∂x/∂c))
 A: You can have --- barring a few technical details --- the Chain Rule:
$$\frac{dy}{dc}=\frac{dy}{dx}\cdot \frac{dx}{dc}.$$
This is because we have that $y=y(c)$ and $x=x(c)$. Consider all of the points on the plane $(x(c),y(c))$ as $c$ runs over its domain. This makes up a curve in the $x$-$y$ plane. 
Assuming that $x(c)$ and $y(c)$ are sufficiently nice, if you zoom close to a point $(x_0,y_0)$ such that $\displaystyle \frac{dx}{dc}\neq 0$ then by the Inverse Function Theorem, then locally, near $x=x_0$, the curve looks like the graph of a function and we can write that $y=y(x)=y(x(c))$.
Now differentiate both sides with respect to $x$, noting that we need a Chain Rule:
$$\begin{align}
\frac{dy}{dc}&=\frac{dy}{dx}\cdot \frac{dx}{dc}
\\ \Rightarrow \frac{dy}{dx}&=\frac{\frac{dy}{dc}}{\frac{dx}{dc}}
\end{align},$$
with the last division OK because we have $\displaystyle \frac{dx}{dc}\neq 0$ near $(x_0,y_0)$ if we assume nice things about $x(c)$ and $y(c)$.
Now if we are on the surface $(x,y,z)=(x,y,f(x,y))$ then $\displaystyle \frac{dy}{dx}$ is the same as $\displaystyle \frac{\partial y}{\partial x}$ on a level curve $z=k$; i.e. $z$ is kept constant.
