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$y=\dfrac{x^2-x+c}{x^2+x+2c} \\ yx^2+yx+2cy=x^2-x+c \\ x^2(y-1)+x(y+1)+2cy-c=0$

As $x$ is real, discriminant of the above equation must be more than or equal to $0$ .

$(y+1)^2-4(y-1)(2cy-c)≥0 \\ y^2(1-8c)+2(1+6c)y+1-4c≥0$

Now the above must be true $\forall y\in \mathbb{R}$ , so the discriminant of this equation must be $≤0$ and $1-8c>0\Rightarrow c<\frac{1}{8}$ .

$2^2(1+6c)^2-4(1-8c)(1-4c)≤0 \\ 4c(c+6)≤0 \Rightarrow c\in [-6,0]$

Now, we have to check the boundary points, at $c=-6$ , $y=\frac{(x-3)(x+2)}{(x-3)(x+4)}\Rightarrow y= \frac{x+2}{x+4}, x≠3$

Therefore, $\frac{3+2}{3+4}=\frac{5}{7}$ will not be included in the range of $y$ as it is only possible for $x=3$ but $x$ should not be equal to $3$ , so $c=-6$ should not be included in our answer.

Similarly, checking for $c=0$ ,

$y=\frac{x^2-x}{x^2+x}\Rightarrow y=\frac{x-1}{x+1},x≠0$

Therefore, $\frac{0-1}{0+1}=-1$ will not be included in the range of $y$ as it is possible only for $x=0$ but $x$ should not be equal to $0$ , so $c=0$ will also not be included in our answer.

$\therefore c\in (-6,0) \\ \Rightarrow -6+0=\boxed{-6}$