Derivative of an Implicit Function

Let two variables x and y are related to a given implicit relation f(x,y) =0. In general, this defines y as a multi-valued function of x. However, the simplest way of dealing with such a function is to introduce a third variable z as under:

Z = f(x, y)

However, z = 0 for all x and y related by f(x,y) =0. Thus the given implicit function can be studied by relating those values of x and y which makes z equal to zero in a explicit function z = f(x,y).

As x and y vary in any way, independently or not, the complete differential of z is given by

Dz = f_x dx + f_y dy

If, (x,y) are values showing z =0 and if, dx and dy are variations from these values keeping z = 0, then z = 0.

Thus, f_x dx + f_y dy = 0

Hence,  dy/dx = fx/fy