3a/(a-b)(a+b) - 3/(b-a)

I'm assuming that we have :

3a/[(a-b)(a+b)] - 3/(b-a)     note... -3/(b-a)  =  -3 / [- (a - b) ] =  + 3/(a - b)     the negative signs "cancel"

This gives us

3a/[(a-b)(a+b)] + 3/(a-b)       multiply the second fracton by (a + b) on top and bottom

3a/[(a-b)(a+b)]  + 3(a + b)/ [(a - b) (a + b)]=   {combine everything over a common denominator}

[3a + 3(a + b)] / [(a - b)(a + b) ] =

[3a + 3a + 3b ] /   [(a - b)(a + b) ] =

[6a + 3b] / [a^2 - b^2] =

3[2a + b] / (a^2 - b^2)