We have f(x) = (3x + 1)*e^-x

We use the product rule to find f'(x)

f'(x) = (3x + 1)'*e^-x + (3x + 1)*(e^-x)'

=> 3*e^-x - (3x +1)e^-x

=> 3*e^-x - f(x)

f''(x) = -3e^-x - f'(x)

=> -3e^-x - 3e^-x + f(x)

Now f(x) + 2f'(x) + f''(x)

=> f(x) + 2(3*e^-x - f(x)) + (-3e^-x - 3e^-x + f(x))

=> f(x) + 6e^-x - 2f(x) - 2*3e^-x + f(x)

cancel the common terms

=> 0

As f(x) + 2f'(x) + f''(x) = 0

=> f(x) = -2f'(x) - f''(x)

**This proves that f(x) = -2f'(x) - f''(x)**

To verify the identity, we'll have to calculate the first and the 2nd derivatives of the fucntion f(x).

f'(x) = [(3x+1)*e^-x]' = 3e^-x - (3x+1)*e^-x

f"(x) = [f'(x)]' = [3e^-x - (3x+1)*e^-x]'

f"(x) = -3e^-x - 3e^-x + (3x+1)*e^-x

Now, we'll substitute the expressions of f'(x) and f"(x) into the given identity:

f(x)=-2f'(x)-f"(x)

(3x+1)*e^-x = -2[3e^-x - (3x+1)*e^-x] - [ -3e^-x - 3e^-x + (3x+1)*e^-x]

We'll remove the brackets:

(3x+1)*e^-x = -6e^-x + 2(3x+1)*e^-x + 3e^-x + 3e^-x - (3x+1)*e^-x

We'll combine and eliminate like terms:

(3x+1)*e^-x = 2(3x+1)*e^-x - (3x+1)*e^-x

**(3x+1)*e^-x = (3x+1)*e^-x q.e.d.**