If you want to get into the technicalities of this, start with wikipedia. 
If you have a manageable number of data points plotted on graph paper, draw a couple of trial straight lines that look like they may be good fits. One at a time, for each line, measure and add up the "errors", i.e., the discrepancy between each plotted data mark and the corresponding point on the straight line fit. Finally, whichever straight line has the least total "errors" is the best fit!
In practice, mathematicians usually go for a more complicated way to evaluate the "errors" in the fit. (Well, they wouldn't be mathematicians if they did otherwise!) 
Plenty of free online resources will calculate the line of best ft for you, if you provide your set of data. I like wolfram alpha, it can do almost anythng! Check it out when you have a spare hour (or six): http://www.wolframalpha.com
+130511 I see you have made a start on this one
You have the first derivative
2 / [x^3 √(2/x^2 - 1/x^4)]^(1/2) which we can rewrite as
2 / [x^3 √(2x^-2 - x^-4)] =
(2x^-3)(2x^-2 - x^-4)^(-1/2)
Note that this derivative is never 0 because there s nothing in the numerator that would make it = 0
The second deivative is
(-6x^-4)(2x^-2 - x^-4)^(-1/2) + (2x^-3)(-1/2)(2x^-2 - x^-4)^(-3/2)(-4x^-3 + 4x^-5) =
(-6x^-4)(2x^-2 - x^-4)^(-1/2) + (2x^-3)(2x^-2 - x^-4)^(-3/2)(2x^-3 - 2x^-5) factor this
[(2x^-4)(2x^-2 - x^-4)^(-3/2)] [ -3(2x^-2 - x^-4) + (x)(2x^-3 - 2x^-5)] =
[(2x^-4)(2x^-2 - x^-4)^(-3/2)][-6x^-2 + 3x^-4 + 2x^-2 - 2x^-4] =
[(2x^-4)(2x^-2 - x^-4)^(-3/2)] [x^-4 - 4x^-2]
We need to look at this diffcult one with several separate graphs (this is the only way I see this....)
Here's the graph of sin-1(1 - 1/x^2)...............https://www.desmos.com/calculator/lgofrknb1d
Notice that this graph has the same domain as the derivative........and notice that it is concave down at all points where it is defined
Now......let's look at the second derivative in pieces
Here's the graph of the first two functions .........https://www.desmos.com/calculator/bqzeqzo4x4
And notice that these graphs are positive for all values in the domain
But.....let's look at the graph of the last function.........https://www.desmos.com/calculator/ekkinenwjq
Notice that this graph is negative for all values in the domain.......and putting this together with the first two functions, which are both positive, we see that the curve is concave downward at all points because...... (first two functions positive) x (last function negative) = negative
And there aren't any critical points to evaluate...the curve is always concave down, but there aren't any "max's".....the only "critical" points are ±1√/2......but at these points...the graph has a vertical tangent....
I think this is correct.......let me know if you have questions.....I did the best I could with this one !!!
