1)

F (x) = [3 x-2 (-x)]/[3 x+2 (-x)] The denominator of the numerator is the same as the denominator multiplied by 2 x:

f(x)=(6^x- 1)/(6^x+ 1)

Because: 6 x > 0 is constant.

So: denominator 6 x+1>; 0 is constant.

So the domain of f(x) is the real number range r, which is symmetrical about the origin.

F (-x) = [6 (-x)-1)]/[6 (-x)+1} The numerator denominator is the same as multiplying by 6 x:

=( 1-6^x)/( 1+6^x)

=-f(x)

So: f(x) is odd function.

2)

f(x)=(6^x- 1)/(6^x+ 1)=(6^x+ 1-2)/(6^x+ 1)= 1-2/(6^x+ 1)

Because: 6 x is a monotonically increasing function on R.

So: 6 x+ 1 is a monotonically increasing function on R.

So: -2/(6 x+ 1) is a monotonically increasing function on R.

So: f(x) is a monotonically increasing function on R.

3)

Because: 6 x+ 1 > 1

So: 0

So: -2

So: 1-2

So the range of f(x) is (-1, 1).