Answer
$dy/dx=\frac{(1+x^2)^2(sec(x-y))^2+2xy}{(1+x^2)^2(sec(x-y))^2+x^2+1}$
Work Step by Step
Take the derivative as is on either side of the equation:
$(sec(x-y))^2\times(1-dy/dx)=\frac{(dy/dx)x^2+dy/dx-2xy}{(1+x^2)^2}$
Move all terms with dy/dx onto one side of the equal sign and distribute the dy/dx out of each term:
$dy/dx((1+x^2)^2(sec(x-y))^2+x^2+1)=(1+x^2)^2(sec(x-y))^2+2xy$
Isolate dy/dx by dividing both sides by the terms:
$dy/dx=\frac{(1+x^2)^2(sec(x-y))^2+2xy}{(1+x^2)^2(sec(x-y))^2+x^2+1}$