Find dy/dx by implicit differentiation.

$\displaystyle tan(x-y) = \frac {y}{(3 + x^2)}$

My Steps:

$\displaystyle \frac {d}{dx} tan(x-y) = sec^2(x-y)(1-y')$

$\displaystyle \frac {d}{dx} \frac {y}{(3 + x^2)} = \frac {(3+x^2)y' - y(2x)}{(3+x^2)^2}$

Therefore:

$\displaystyle sec^2(x-y)(1-y') = \frac {(3+x^2)y' - y(2x)}{(3+x^2)^2}$

Now I know that I need to solve for y' but no matter how long I stare at this problem I can't figure it out.