Answer
\[y^{'}=\frac{(1+x^2)^2\sec^2(x-y)+2xy}{(1+x^2)+(1+x^2)^2\sec^2(x-y)}\]
Work Step by Step
$\tan (x-y)=\large\frac{y}{1+x^2}$ ___(1)
Differentiating (1) implicitly with respect to $x$
$\sec^2 (x-y)\:[1-y^{'}]=\large\frac{y^{'}(1+x^2)-y(1+x^2)^{'}}{(1+x^2)^2}$
$\sec^2 (x-y)-y^{'}\sec^2 (x-y)=\large\frac{y^{'}}{1+x^2}-\frac{2yx}{(1+x^2)^2}$
$\sec^2 (x-y)+\frac{2xy}{(1+x^2)^2}=y^{'}\left[\frac{1}{1+x^2}+\sec^2 (x-y)\right]$
$y^{'}=\Large\frac{\frac{(1+x^2)^2\sec^2(x-y)+2xy}{(1+x^2)^2}}{\frac{1+(1+x^2)\sec^2(x-y)}{(1+x^2)}}$
\[y^{'}=\frac{(1+x^2)^2\sec^2(x-y)+2xy}{(1+x^2)+(1+x^2)^2\sec^2(x-y)}\]
Hence $y^{'}=\Large\frac{(1+x^2)^2\sec^2(x-y)+2xy}{(1+x^2)+(1+x^2)^2\sec^2(x-y)}$.
