Find the value of $\frac{a^2}{a^4+a^2+1}$ if $\frac{a}{a^2+a+1}=\frac{1}{6}$
From the first equation (inverted),
$$\frac{a^2+a+1}a=6$$ or $$\frac{a^2+1}a=5.$$
Then squaring,
$$\frac{a^4+2a^2+1}{a^2}=25$$ or $$\frac{a^4+a^2+1}{a^2}=24.$$
Hint. From the equation, one easily gets $$ a^2=5a-1, \quad a^4=(5a-1)^2=25a^2-10a+1=115a-24 $$ giving in the first expression
$$ \frac{a^2}{a^4+a^2+1}=\frac{5a-1}{120a-24}=\frac{1 \times\color{red}{(5a-1)}}{24\times\color{red}{(5a-1)}}=\frac1{24}. $$
You are asked to express $\dfrac1B=\dfrac1{a^2+1+a^{-2}}$ in terms of $\dfrac1A=\dfrac1{a+1+a^{-1}}$.
Squaring "to see",
$$A^2=(a+1+a^{-1})^2=a^2+1+a^{-2}+2a+2+2a^{-1}=B+2A.$$
This gives us
$$B=A^2-2A=6^2-2\cdot6=24.$$