The given expression expands to $\frac{1}{b^2} - a^2$.  Thus we want $b$ to have the least possible magnitude and $a$ to also have the least possible magnitude.  Our maximum value is therefore $\frac{1}{3^2} - (-2)^2 = \boxed{-\frac{35}{9}}$.
Question:
If $-6\leq a \leq -2$ and $3 \leq b \leq 5$, what is the greatest possible value of $\displaystyle\left(a+\frac{1}{b}\right)\left(\frac{1}{b}-a\right) $?  Express your answer as a common fraction.
Answer:
Let's think step by step.

user
-\frac{35}{9}
If $-6\leq a \leq -2$ and $3 \leq b \leq 5$, what is the greatest possible value of $\displaystyle\left(a+\frac{1}{b}\right)\left(\frac{1}{b}-a\right) $?  Express your answer as a common fraction.
test