Step 1: Set the function equal to $y$.
Write $y=\dfrac{x^2+x+1}{x^2+kx+1}$. For the function to take every real value, this must have a real $x$ for each chosen $y$.
Step 2: Cross multiply into a quadratic in $x$.
This gives $(y-1)x^2+(yk-1)x+(y-1)=0$. We need real $x$, so its discriminant must be $\ge 0$.
Step 3: Form the discriminant.
The condition is $(yk-1)^2-4(y-1)^2\ge0$. Expanding gives $y^2(k^2-4)+y(8-2k)-3\ge0$.
Step 4: Demand it holds for all $y$.
For this quadratic in $y$ to be non-negative for every real $y$, we need the leading part positive, $k^2-4>0$, and its own discriminant $\le0$.
Step 5: Check that discriminant.
It works out to $(8-2k)^2+12(k^2-4)=16(k-1)^2$. For this to be $\le0$ we must have $k=1$.
Step 6: Test the leftover condition.
But at $k=1$, $k^2-4=-3<0$, which breaks the earlier need $k^2-4>0$. So no value of $k$ works, and the answer is the empty set. \[ \boxed{\phi} \]