If 'x' is real then $\frac{x^2-x+c}{x^2+x+2c}$ can take all real values if? the question asks for the interval in which c lies so that $\frac{x^2-x+c}{x^2+x+2c}$ gives all real values for all x belongs to R.
how to proceed in this problem?

Let $y = \frac{x^2-x+c}{x^2+x+2c} $ be a value taken by the given fraction. This gives 
\begin{equation*}
x^2(y-1) +x(y+1) +c(2y-1) = 0
\end{equation*}
should have a real root. Thus 
\begin{equation*}
(y+1)^2 -4c(y-1)(2y-1) \geq 0
\end{equation*}
and hence
\begin{equation*}
y^2(1-8c) + 2y(1+6c)+(1-4c) \geq 0
\end{equation*}
If the given expression takes all values, then the above should be true for all $y$. Thus we have 
\begin{equation*}
(1+6c)^2 - (1-8c)(1-4c) \leq 0
\end{equation*}
and $8c \leq 1$. The above expression simplifies to $24c+4c^2 \leq 0$ and hence $-6 \leq c \leq 0$. 
when $c=-6$, the graph is: Clearly, $y=1$ is an asymptote and hence the expression misses the value 1. Again, the same phenomenon is observed when $c=0$. Thus the required range is $-6 < c < 0$
For negative values outside the range, the graph is  and for positive values of $c$, the graph is  and for values within the range, the graph is 
