How do I find the final concentration after an additional dilution? I took poor notes and didn't record what the final concentration of an amylase solution I  made last year. I recorded that I took $\rm 0.2\ g/10\ ml$ of the amylase solution, and then took $\rm 1\ ml$  of that and diluted to $35 ml$. 

How would I solve for the final concentration?

Would I use :

$$c_1\ V_1 = c_2\ V_2$$

$$\rm(0.2\ g/10ml) \cdot (1\ ml) = (\mathit{c_2}) * ( \mathrm{35\ ml})$$

$$c_2\rm = \frac{0.02\ g}{35\ ml} $$

Therefore, I would take $\rm 0.02 g$ of amalyase powder and add up to $\rm 35\ ml$  with distilled $\ce{H2O}$ to achieve the final concentration that I would have previously had after dilution?
Check your math: It looks like you forgot to divide by $ 10\ \mathrm{ml} $ in finding $ c_2 $. 

Also: the process you are describing in the first paragraph will not produced the same concentration that you started with. If you want $ 35\ \mathrm{ml} $ at the same concentration ($ 0.02\ \mathrm{g/ml} $) by taking some fraction of your original solution and diluting it, you will need to add more solute.

e.g.
$$ \rm(0.02\ g/ml) \cdot (35\ ml) = 0.07\ g $$
$$ \rm(0.07\ g + \mathit{m}) / 35\ ml = (0.02\ g/ml) $$
$$ m \rm = (0.02\ g/ml) \cdot (35\ ml) - 0.07\ g $$

The first expression is the concentration of your fraction (which doesn't change from the original solution, assuming a homogenous solution).

The second expression is equating the concentration that you want (conc. of the original solution, for example, with the concentration of your diluted fraction with $ m\ \mathrm{g} $ of solute added (using the mass we just found).

Then I rearranged this to express the mass of solute needed to obtain this concentration.