How to solve this derivative?

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

F '(x) = (-2x^2 - 2x + x^2 + 2x - 3)/(x^2 + 2x - 3)^2

F '(x) = (-x^2 - 3)/(x^2 + 2x - 3)^2

Find the derivative of the following via implicit differentiation:

d/dx(F(x)) = d/dx(x/(x^2+2 x-3))

The derivative of F(x) is F'(x):

F'(x) = d/dx(x/(-3+2 x+x^2))

Use the quotient rule, d/dx(u/v) = (v ( du)/( dx)-u ( dv)/( dx))/v^2, where u = x and v = x^2+2 x-3:

F'(x) = ((x^2+2 x-3) d/dx(x)-x d/dx(-3+2 x+x^2))/(x^2+2 x-3)^2

The derivative of x is 1:

F'(x) = (-(x (d/dx(-3+2 x+x^2)))+1 (-3+2 x+x^2))/(-3+2 x+x^2)^2

Simplify the expression:

F'(x) = (-3+2 x+x^2-x (d/dx(-3+2 x+x^2)))/(-3+2 x+x^2)^2

Differentiate the sum term by term and factor out constants:

F'(x) = (-3+2 x+x^2-d/dx(-3)+2 d/dx(x)+d/dx(x^2) x)/(-3+2 x+x^2)^2

The derivative of -3 is zero:

F'(x) = (-3+2 x+x^2-x (2 (d/dx(x))+d/dx(x^2)+0))/(-3+2 x+x^2)^2

Simplify the expression:

F'(x) = (-3+2 x+x^2-x (2 (d/dx(x))+d/dx(x^2)))/(-3+2 x+x^2)^2

The derivative of x is 1:

F'(x) = (-3+2 x+x^2-x (d/dx(x^2)+1 2))/(-3+2 x+x^2)^2

Use the power rule, d/dx(x^n) = n x^(n-1), where n = 2: d/dx(x^2) = 2 x:

F'(x) = (-3+2 x+x^2-x (2+2 x))/(-3+2 x+x^2)^2

Expand the left hand side to find the answer:

F'(x) = (-3+2 x+x^2-x (2+2 x))/(-3+2 x+x^2)^2

y = u/v where both u and v are functions of x

dy/dx = [v*(du/dx) - u*(dv/dx)]/v^2

F'(x) = [(x^2 + 2x - 3) - x(2x + 2)]/[(x^2 + 2x + 3)^2]

F'(x) = [-x^2- 3]/[(x^2 + 2x + 3)^2]