How does one show that the polynomial system $F(x) = 0,$ where $F:\mathbb{C}^n \rightarrow \mathbb{C}^n,$ has only isolated roots?

I'm a bit late to the party, but here are five answers, not using Groebner basis: (1), if you can show that the zero set of $F$ is local complete intersection, then the local dimension must all be zero, i.e., the zero sets are geometrically isolated sets of points. But the converse is not true. (2), if you weaken your problem a little bit by replacing $\mathbb{C}^n$ by $(\mathbb{C}^\ast)^n=(\mathbb{C}\setminus\{0\})^n$, then you can use Bershtein's second theorem, which states that $F$ has only isolated solutions in $(\mathbb{C}^\ast)^n$ if and only if all "initial systems" of $F$ has no solutions in $(\mathbb{C}^\ast)^n$. Indeed, we can even know the precise number of solutions counting multiplicity: it is the "mixed volume" of the Newton polytope of $F$ (commonly known as the "BKK bound"). (3) For your original question, over $\mathbb{C}^n$, there is a generalization of (2) by Li & Wang: which, basically, states that if you add 0 constant terms to F, i.e., add 0 to each polynomial in $F$ (which does not change $F$ itself but alters its Newton polytope), then (2) will give you the desired answer over $\mathbb{C}^n$. (4) Both (2) and (3) has been "translated" into the more fashionable (for now) language of tropical geometry. So you basically have to verify some rather technical conditions for all induced "pretropism". (5) Complex polynomials are actually analytic functions, so the zero set of $F$ is actually a holomorphic variety. The local parametrization theorem can actually tell you the condition under which you only have isolated solutions. This is certainly an overkill, but it is much more general. 

