How to find that two circles form a knot? Suppose $f$ and $g$ are two embeddings of $S^1$ in $R^3$ or in $S^3$. How do I show whether they form a link or not?
 A: I think the common interpretation of your question is to determine whether you can pull the components apart with an ambient isotopy. This is a harder problem than you might expect, but it is doable. The key is that being able to pull the knots apart is equivalent to whether there is a topological sphere in the complement in $S^3$ which is not contractible. (A really large sphere containing everything is contractible in $S^3$, but not $\mathbb{R}^3$.) Normal surface theory lets us check for interesting $S^2$s algorithmically. 
You can triangulate the complement of the link (already not a trivial operation) and get a finite-dimensional space of normal surfaces. Then you can construct a polytope such that if there is any incompressible $S^2$, there is one which is a vertex of the polytope. So, you find all of the vertices and check them. 
A: Gauss's linking integral will give the linking number:
$$N = \frac{1}{4\pi}\int\int \frac{\vec{f}(\sigma)-\vec{g}(\tau)}{|\vec{f}(\sigma)-\vec{g}(\tau)|^3}\cdot \left(\frac{d\vec{f}}{d\sigma}\times \frac{d\vec{g}}{d\tau}\right)\;d\sigma\;d\tau$$
Per comment by Alon Amit, the linking number can be zero for maps that can't be teased apart such as the Whitehead link, but if it's nonzero, they definitely can't be separated. (And by the way, even an "unlink" is a "link" so I'm not certain this is what you're looking for.)
