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

$$B=A^2-2A=6^2-2\cdot6=24.$$