Some curves are described by equations of the form F(x,y)=0. For example, x2+y2=1 is the equation of circle of radius 1. For the upper semi-circle we can solved it for y and write y=1x2 and for the lower semi-circle y=1x2. There is no function y=f(x) near (±1,0) that  satisfies F(x,f(x))=0.

Let F(x,y)=0. Suppose F(x0,y0)=0 and F has continuous first partial derivatives and so is differentiable. There are two questions that we try to  answer:

  1. Can we solve for y as a function of x near (x0,y0)? In other words, can we find a function y=f(x) defined on some interval I=(x0h,x0+h) (for h>0) such that F(x,f(x))=0?
  2. If such a function exists, what is f(x0)=dydx|x=x0?

Suppose such a function exists. To find f(x0)

Method (a): We use the chain rule:
dFdx=Fxdxdx=1+Fydydx=f(x)=0. so that

f(x0)=dydx|x=x0=Fx(x0,y0)Fy(x0,y0),providedFy(x0,y0)0.

Method (b): Because F is constant and therefore its the total differential is zero.
dF=Fxdx+Fydy=0. If we divide the above equation by dx and assume Fy(x0,y0)0, we obtain the same result as method (a).

We assumed that a function y=f(x) existed and then showed the condition Fy(x0,y0)0 was required for calculation of f(x0). In fact, it can be proved this condition, Fy(x0,y0)0, is sufficient for the existence of y=f(x) with the aforementioned conditions. The condition Fy(x0,y0)0 means that the tangent line to the level curve F(x,y)=0 is not vertical, and therefore, a part of the level curve — close enough to the point (x0,y0)— can be the graph of the function y=f(x). When Fy(x0,y0)=0, you may not be able to find such a function. For example in Fig. 1, if we just keep the shaded disk around (x0,y0) and remove the rest of the level curve, what we get can be the graph of a function, because any vertical line now does not intersect this part of the curve more than once. However, at A or B, where the tangent line is vertical, we cannot find a disk around them (to keep the curve and remove the rest) where a vertical line does not intersect the level curve twice. Therefore the level curve near A or B cannot be the graph of a function.

Figure 1.

Noting Fx and Fy are both functions of x and y, higher derivatives of y with respect to x can be found by successive differentiation with respect to x provided higher partial derivatives of F(x,y) exist:
Fx+Fyy=02Fx2+22Fxyy+2Fy2y2+Fyy=0 3Fx3+33Fx2yy+33Fxy2y2+3Fy3y3+32Fy2yy+32Fxyy+Fyy=0

Example 1
Given y3x33y2x2+6=0, find dy/dx.

Solution
Let F(x,y)=y3x33y2x2+6. Thus:
Fx(x,y)=3x22x,Fy(x,y)=3y26y, and it follows that
dydx=FxFy=3x22x3y26y. We cannot use the above formula when 3y26y=0.
3y26y=3y(y2)=0y=0 or y=2. As you can see in Fig. 2, when y=0 and y=2, the tangent line to the curve is vertical.

Figure 2.

Now suppose the equation F(x,y,z)=0 is given, where F has continuous partial derivatives. If F(x0,y0,z0)=0 and F/z(x0,y0,z0)0, z near (x0,y0,z0) can be written as a function of x and y, namely z=f(x,y). In other words, the level surface F(x,y,z)=0 can be locally the graph of a function z=f(x,y). To find the partial derivatives of f, we differentiate the equation F(x,y,z)=0 with respect to x and y:
Fx+Fzzx=0,Fy+Fzzy=0. Therefore:

zx(x0,y0)=fx(x0,y0)=Fx(x0,y0,z0)Fz(x0,y0,z0) and
zy(x0,y0)=fy(x0,y0)=Fy(x0,y0,z0)Fz(x0,y0,z0).

Example 2
Given z2x2y2=3, find zx and zy.

Solution
Let F(x,y,z)=z2x2y23. Then Fx=2x,Fy=2y,Fz=2z. Therefore:
zx=FxFz=2x2z=xz,zy=FyFz=2y2z=yz. Note that Fz=2z0 and the above expressions are valid everywhere because for any point on the level surface, |z|3 (note z2=3+x2+y23).

In general we have the following theorem.

Theorem 1. (Implicit Function Theorem) Suppose F(x1,,xn,z)=0 and F is of class C1 (i.e., has continuous first partial derivatives). Assume:
F(x0,z0)=0andFz(x0,z0)0, where x0Rn and z0R. Then there is a neighborhood U of x0 in Rn, a neighborhood V of z0 in R, and a function f:URnV of class C1 such that if x=(x1,,xn)U and zV satisfy F(x,z)=0, then z=f(x). The partial derivatives of f are given by:
fxi(x0)=Fxi(x0,z0)Fz(x0,z0).