You need to use quotient rule to evaluate the derivative of the given function, such that:

`f'(x) = ((x - 3)'(x^2 + 2) - (x - 3)(x^2 + 2)')/((x^2 + 2)^2)`

`f'(x) = (1*(x^2 + 2) - (x - 3)(2x))/((x^2 + 2)^2)`

`f'(x) = (x^2 + 2 - 2x^2 + 6x)/((x^2 + 2)^2)`

`f'(x) = (-x^2 + 6x + 2)/((x^2 + 2)^2)`

**Hence, evaluating the derivative of teh function, using the quotient rule, yields **`f'(x) = (-x^2 + 6x + 2)/((x^2 + 2)^2).`

We'll use the limit method to calculate the value of derivative of a function in a given point.

lim [f(x) - f(1)]/(x-1) = lim [(x-3)/(x^2+2) + 2/3]/(x-1)

lim [(x-3)/(x^2+2) + 2/3]/(x-1) = lim (3x-9+2x^2+4)/(x-1)

lim (3x-9+2x^2+4)/(x-1) = lim (3x-5+2x^2)/(x-1)

We'll substitute x by 1:

lim (3x-5+2x^2)/(x-1) = (3-5+2)/(1-1) = 0/0

Since we've obtained an indeterminacy, we'll apply L'Hospital rule:

lim (3x-5+2x^2)/(x-1) = lim (3x-5+2x^2)'/(x-1)'

lim (3x-5+2x^2)'/(x-1)' = lim (3+4x)/1

We'll substitute x by 1:

lim (3+4x)/1 = (3+4)/1

f'(1) = lim (3x-5+2x^2)/(x-1) = 7

f'(1) = 7