Mohammad, 

flatness is **not** simple, so you are not going to get a simple overall definition. On the other hand, in the example you mention in your comment, there is a simple criterion:

> Let $f:X\to Y$ be a morphism of schemes such that $X$ is reduced and irreducible (most likely satisfied in the cases you are interested in, at least for now) and $Y$ is a smooth curve. Then if $X$ dominates $Y$ (i.e., the image is dense in $Y$) then $f$ is flat. 

In particular $f:\mathbb C^2\to \mathbb C$ given by $(x,y)\mapsto xy$ is flat.

Over higher dimensional basis it is a little more complicated, but there is one simple criterion you can check:

> If $f$ is flat, then the fibers have the same dimension.

For example a blowup is the typical not flat map (think of the meaning of "flat"). 

On the other hand, the fibers being equidimensional is certainly not enough for a morphism to be flat. An example is provided by taking $X$ to be the union of two planes meeting in a single point mapping to just one copy of the plane in the obvious way. All fibers are $0$-dimensional, but the morphism is not flat.

However, there is a criterion that is pretty useful and says that if both $X$ and $Y$ are relatively nice, then flatness is equivalent to that. 

> Let $f:X\to Y$ be a dominant morphism of irreducible varieties such that $X$ is Cohen-Macaulay and $Y$ is regular (smooth if you prefer), then $f$ is flat if and only if the fibers are equidimensional. 

Finally, my suggestion is that in order to get a better feeling for flatness, try looking at finite morphisms as in the above example. If you understand those, you essentially have understood all flat maps.