Rewrite
x^2/(2x^2+7x+6) = 1/2(1 - (7x+6)/(2x^2+7x+6))
Factor 2x^2+7x+6 as (2x+3)(x+2). Perform a partial fraction decomposition to give
x^2/(2x^2+7x+6) = 1/2 (1 - 8/(x+2) + 9/(2x+3))
Consider the nth derivative of (x+a)^(-1).
"d"/("d"x) (x+a)^(-1) = (-1) * (x+a)^(-2)
"d"^2/("d"x^2) (x+a)^(-1) = (-1) * (-2) *(x+a)^(-3).
Easy to guess that "d"^n/("d"x^n) (x+a)^(-n)= (-1)^(n)n!(x+a)^(-(n+1)).
Proceed by induction. Assume for n=k that "d"^k/("d"x^k) (x+a)^(-k) = (-1)^(k)k!(x+a)^(-(k+1)).
Then,
"d"^(k+1)/("d"x^(k+1)) (x+a)^(-k) = "d"/("d"x) ("d"^k/("d"x^k) (x+a)^(-k)),
"d"^(k+1)/("d"x^(k+1)) (x+a)^(-k) = "d"/("d"x) (-1)^(k)k!(x+a)^(-(k+1)),
"d"^(k+1)/("d"x^(k+1)) (x+a)^(-k) = (-1)^(k)k! (-(k+1)) (x+a)^(-(k+2)),
"d"^(k+1)/("d"x^(k+1)) (x+a)^(-k) = (-1)^(k+1)(k+1)! (x+a)^(-((k+1)+1)),
Then if the expression holds for n=k this implies it holds for n=k+1. It is true for n=1 thus it is true for all positive integer n by mathematical induction.
"d"^n/("d"x^n) x^2/(2x^2+7x+6) = 1/2 "d"^(n)/("d"x^n) (1 - 8/(x+2) + 9/(2x+3))
"d"^n/("d"x^n) x^2/(2x^2+7x+6) = 1/2 "d"^(n)/("d"x^n) (1 - 8/(x+2) + (9/2)/(x+3/2))
"d"^n/("d"x^n) x^2/(2x^2+7x+6) = "d"^(n)/("d"x^n) (2) - 4 * "d"^(n)/("d"x^n) (x+2)^(-1) + 9/4 * "d"^(n)/("d"x^n) (x+3/2)^(-1)
Our expression will hold for positive integer n. Clearly "d"^n/("d"x^n) 2 = 0 for n>=1.
Then, substituting the result proved earlier,
"d"^n/("d"x^n) x^2/(2x^2+7x+6) = 9/2^2(-1)^(n)n!(x+3/2)^(-(n+1)) - 4(-1)^(n)n!(x+2)^(-(n+1)),
"d"^n/("d"x^n) x^2/(2x^2+7x+6) = (-1)^(n)(n!)(9/(2^2(x+3/2)^(n+1)) - 4/((x+2)^(n+1))),
"d"^n/("d"x^n) x^2/(2x^2+7x+6) = (-1)^(n)(n!)((9*2^(n-1))/((2x+3)^(n+1)) - 4/((x+2)^(n+1))).
