9.2. It is known that $\frac{a^{2} b^{2}}{a^{4}-2 b^{4}}=1$. Find all possible values of the expression $\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$.
Answer: $\frac{1}{3}$.

Solution. Multiply by the denominator and transform the equality in the condition to $b^{2}\left(a^{2}+b^{2}\right)=a^{4}-b^{4}=\left(a^{2}+b^{2}\right)\left(a^{2}-b^{2}\right)$. Here, $a, b$ cannot both be zero at the same time, otherwise the denominator of the fraction in the condition would turn to 0. Therefore, the bracket $a^{2}+b^{2}$ is also not equal to zero. Dividing by it, we get $b^{2}=a^{2}-b^{2}$, from which $a^{2}=2 b^{2}$ (where $a, b$ are not equal to 0). Substituting this expression into the fraction $\frac{a^{2}-b^{2}}{a^{2}+b^{2}}$, we find that it equals $\frac{b^{2}}{3 b^{2}}=\frac{1}{3}$.

Grading criteria. No points are deducted for dividing by an expression that could be zero: minus 2 points.

If solved by dividing by $b^{2}$ without considering $b=0$, minus 2 points; if the expression is transformed to express $a^{2}+1$ point.