find dy/dx for tan(x - y) = (xy) / (x^2 + y^2)

[bjornski7]

find dy/dx for tan(x-y)=(xy)/(x^2+y^2)

So far I have

sec^2(x-y)(1-dy/dx) = ((y+xdy/dx)(x^2+y^2)-(xy)(2x+2ydy/dx))/((x^2+y^2)^2)

but i don't know how to get dy/dx from that
thank you,
blake

 

[bjornski7]

Re: dy/dx for tan(x-y)=(xy)/(x^2+y^2)

aha, i use ln