The only thing you can do here is to try to simplify, so I'll assume that's what you want to do.
[tex]\dfrac{b^2-a^2}{2a^2+ab-3b^2}[/tex]
The numerator is a difference of squares, so it's easy to factorize:
[tex]b^2-a^2=(b-a)(b+a)[/tex]
In the denominator, we factorize by grouping:
[tex]2a^2+ab-3b^2=2a^2-2ab+3ab-3b^2=2a(a-b)+3b(a-b)=(2a+3b)(a-b)[/tex]
So we have
[tex]\dfrac{b^2-a^2}{2a^2+ab-3b^2}=\dfrac{(b-a)(b+a)}{(2a+3b)(a-b)}=-\dfrac{b+a}{2a+3b}[/tex]
