what was Pierre de Fermat's (or Issac Barrow's) method for finding
lines tangent to a curve at a given point?  I need to use their method to find
the equation of the line tangent to the curve y=x^3+2x at the point
(1,3).

Hello dingus!
After some searching, I came across these two pages, which I'm
confident they'll be clear enough for you to understand. The first one
is the best.

History of the Differential from the 17th century
http://www.math.wpi.edu/IQP/BVCalcHist/calc2.html
(go to item 2.3)

Isaac Barrow (1630-1677)
http://www.maths.tcd.ie/pub/HistMath/People/Barrow/RouseBall/RB_Barrow.html

As you can see, he used a method of similar triangles; but he was
actually using the concept of derivatives, although he as unable to
formalize it.

As I can only write text here, writing a detailed solution for your
particular exercise is quite difficult here, but I'm sure that with
the first link you'll be able to solve it with no trouble. Anyway,
I'll provide the solution to your exercise (the commenter xarqui got
it right) using the derivatives method, so you can then check if you
got the right answer.

f(x) = x^3 + 2x
f'(x) = 3x^2 + 2
f'(1) = 5

The solution will be then a line with slope=5 that youches (1,3), so:
y = 5x + b
3 = 5*1 + b
b = -2

So, the tangent line you're looking for is y = 5x + 2. You should
arrive to the same result using Fermat's method.

I hope this was clear enough. If you have any further questions,
please don't hesitate to request clarification.


Best wishes,
elmarto

This may seem a silly comment considering the clarity of your
question, but would another method do?
For example just differentiating the equation to yield y' = 3x^2 +2. 
This will tell you that the slope at (1,3) is 5.
From this it follows that the y-intercept of the tangent is -2, and
its equation is:
 y = 5x - 2.

Nice job elmarto, but maybe a typo at the end:  should be y = 5x - 2, no?