Proving differentiable mapping is onto
Solution 1:
1) If $\det f'(x) = 0$, then there is a vector $h$ such that $f'(x)h = 0$. What happens now, from the definition of the derivative?
2) The easiest way I can see to do this is use invariance of domain. Since you've shown $f$ is 1-1, you know that $f(\mathbb{R}^n)$ is an open set. (I suppose you secretly don't need invariance of domain, only the IFT.) Now, I claim you can also show it's closed, using your condition. (Recall some facts about Cauchy sequences for this.) Hence it must be all of $\mathbb{R}^n$.