kx^3 + 3kx^2 + 3kx + k + 1 - 1/x^2 = 0

kx^5 + 3kx^4 + 3kx^3 + (k+1)x^2 - 1 = 0    ** Eqn #1

so the sum of ALL roots is also -3  (-b/a, Vieta's equations)

or factor the first term

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

(x+1)(x-1)/x^2 + k(x + 1)(x^2 - x + 1) = 0

So it is clear that (x+1) is a factor and in fact, by inspection of the original equation, in retrospect, x = -1 is a solution.

If x=-1 happens to also be one of the lowest values, the the other is -2 (may or may not be)

If x = -2 is a solution then

(3/4) + k(-1) = 0, or k = 3/4 (if at least one other solution is > -1)

Substitute k = (3/4) into Eqn #1 to get

(3/4)x^5 + 3(3/4)x^4 + 3(3/4)x^3 + (7/4)x^2 - 1 = 0 

3x^5 + 9x^4 + 9x^3 + 7x^2 - 4 = 0   ** Eqn #2

Divide Eqn #2 by (x+1)

3x^5 + 3x^4 + 6x^4 + 6x^3 + 3x^3 + 3x^2 + 4x^2 + 4x - 4x - 4 = 0

(x+1)(3x^4 + 6x^3 + 3x^2 + 4x - 4) = 0

Now divide 3x^4 + 6x^3 + 3x^2 + 4x - 4  by (x+2)

3x^4 + 6x^3 + 3x^2 + 6x - 2x - 4

(x+2)(3x^3 + 3x - 2)

Eqn #2 becomes (x+1)(x+2)(3x^3 + 3x - 2)

I believe Descartes Rule of Signs says the cubic above has at one positive real root and no negative real roots bringing the total number of real roots to 3, satisfying the final constraint of the problem.

So k = 3/4 checks out as a valid solution

Putting the original equation with k=3/4 into Wolfram Alpha indeed shows 3 real solutions at {-2, -1, approx 0.52334} and 2 complex solutions.

https://www.desmos.com/calculator/ipymt6na7b  shows the graph with 3 real roots