in this calculus. formula used,
(a - b)^2 = a^2 - 2.a.b + b^2
(ab)^2 = (ab) (ab) or a^2.b^2

Explanation

Here,
\large\frac{a}{b}\ =\ \frac{b}{a}\normalsize\ – 2
\rightarrow\large \frac{a}{b}\ -\ \frac{b}{a}\normalsize\ = -2
\rightarrow\large \frac{a^2\ -\ b^2}{ab}\normalsize\ = -2 _[L.C.M]
\rightarrow\large \left(\frac{a^2\ -\ b^2}{ab}\right)^2\normalsize\ = \left(-2\right)^2 _{[add square in both side]}
\rightarrow\large \frac{\left(a^2\ -\ b^2\right)^2}{\left(ab\right)^2}\normalsize\ = 4
\rightarrow\large \frac{\left(a^2\right)^2\ -\ 2.a^2.\ b^2\ +\ \left(b^2\right)^2}{\left(ab\right)\left(ab\right)}\normalsize\ = 4
\rightarrow\ \frac{a^4\ -\ 2.a^2.b^2\ +\ b^4}{a^2b^2}\ =\ 4
\rightarrow\ a^4\ -\ 2.\ a^2.\ b^2\ +\ b^4\ =\ 4a^2b^2 _{[cross multiply]}
\rightarrow\ a^4\ +\ b^4\ =\ 4a^2b^2\ +\ 2a^2b^2
\rightarrow\ a^4\ +\ b^4\ =\ 6a^2b^2
\therefore\ a^4\ +\ b^4\ =\ 6a^2b^2

Now,
\large\frac{a^2}{b^2}\ +\ \frac{b^2}{a^2}
\rightarrow\large \frac{a^4\ +\ b^4}{a^2b^2} _[L.C.M]
\rightarrow\large \frac{6a^2b^2}{a^2b^2} [ put the value of a^4 + b^4 ]
\rightarrow\ 6
Ans: \large\frac{a^2}{b^2}\ +\ \frac{b^2}{a^2}\ =\ 6.