You need to evaluate the derivative of the composed function `h(x) = f(g(x))` , hence, you need to use the chain rule, such that:

`h'(x) = f'(g(x))*g'(x)`

You need to evaluate the derivative of the function `h(x)` at `x = 1` , such that:

`h'(1) = f'(g(1))*g'(1)`

The problem provides the values of `g(1) = 2` and `g'(1) = -2` , hence, you need to replace 2 for g(1) and -2 for `g'(1)` such that:

`h'(1) = f'(2)*(-2)`

The problem also provides the value of `f'(2) = 3` , such that:

`h'(1) = 3*(-2) => h'(1) = -6`

Hence, evaluating the derivative of the function `h(x)` ,  at `x = 1` ,  yields `h'(1) = -6` .

You also need to prove that the derivative of an even function is odd and derivative of an odd function is even.

You need to remember that the odd function follows the rule, such that:

`f(-x) = -f(x)`

Differentiating the function `f(-x)` , using the chain rule, yields:

`(f(-x))' = f'(-x)*(-x)' => (f(-x))' = f'(-x)*(-1)`

Replacing `-f(x)` for `f(-x)` yields:

`(-f(x))' = -f'(-x)` => `f'(x) = f'(-x)` even function

Hence, differentiating the odd function, using the chain rule, yields an even function.