The function is defined because x^2+x+1 always >0.
Further, ln(y) is monotone increasing and therefore it has min and max where y has min and max.
x^2+x+1 has a global minimum at x=-1/2 and it is 3/4, and global maximum at x=1 (it is 3). So f has minimum ln(3/4) and maximum ln(3) .
Take the derivative:
Set the numerator equal to 0.
Find the boundaries
the absolute max is `ln3 ` at `x=1 `
the absolute min is `ln0.75 ` at `x=-1/2 `