Here is the surface defined by z=x^2+y^2, with a square domain [-1,1] x [-1,1] and another surface with the unit disk domain. Even though the defining equation is the same in both cases, the functions are different since they have different domains.

Here is a picture of the saddle: z=x^2-y^2
cut with plane y=0
or cut with plane x=0

Similarly, the saddle:z=2xy
cut with plane x=y or
cut with plane x=-y

Actually these graphs z=x^2-y^2 and z=2xy (over the whole plane) have
exactly the same shape since rotation around the z-axis by 45 degrees
takes one graph into the other.

Contour graphs (curves) are made by intersecting the surface
with the planes z=Constant for various constants.
Here is the Monkey Saddle:
z=y(y^2-x^2)

Here is another nice cubic equation:
z=xy(y^2-1)

The contour lines are drawn on the surface then lifted up
straight up to a plane and then shown flat in small window
on the right hand side. Negative coutour lines are red,
positive blue, and zero are yellow.

Besides polynomials the sine, cosine, and exponential functions
are well behaved.
Here is the surface:
z=sin(x+y)

Notice that by holding x fixed say x=C the curves
z=sin(C+y) are just shifted sine curves.

Here is the graph of the surface:
z=sin(xy) and also the egg carton:
z=sin(x)sin(y)

A rational function is the ratio of two polynomials. In one variable (since a polynomial only has finitely many zeros) a rational function's domain is all of the real line except for finitely many points. The situation is different in two variables, consider z=1/xy , what happens above the x and y axis? Note that when taking a partial derivative the limit is taken along a direction parallel to the x or y axis.

One the other hand there could just be just one bad point at (0,0) z=1/(x^2+y^2) .

Is it true that if the limit as x goes to zero of f(x,0) is 0 and the limit as y goes to zero of f(0,y) is zero, then the limit of f(x,y) as (x,y) goes to (0,0) is 0? ANSWER (another view with eye ball at (6,-2,3) ANSWER ) What is the limit of f(x,x) as x goes to 0?

Well suppose that f(x,y) goes to zero along every ray going to (0,0). Surely that is enough to imply that the limit f(x,y) = 0 as (x,y) goes to (0,0)? ANSWER (another view with eye ball at (2,6,3) ANSWER ) The yellow grid is the surface {(x,y,z): y=x^2}.

Executable and source: progs.zip. These programs are written in TrueBasic for MSDos computers and zipped into an archive using PKZIP (use PKUNZIP) to unzip.