Problem: $\displaystyle x^2+xy-y^2=4$

Solution attempt:$\displaystyle \frac{d}{dx} (x^2)+\frac{d}{dx}(xy)-\frac{d}{dx}(y^2)=\frac{d}{dx}(4)$

$\displaystyle 2x+y+xy'-2yy'=0$

$\displaystyle (2x+y)+(xy'-2yy')=0 $

$\displaystyle y'(x-2y)=-(2x+y)$

$\displaystyle y'=\frac{-(2x+y)}{x-2y}$

$\displaystyle y'=\frac{-2x-y}{-2y+x}$

$\displaystyle y'=\frac{2x-y}{2y+x}$

Answer in book: $\displaystyle y'=\frac{2x+y}{2y-x}$

Notes:I've always had a mundane problem with negatives and fractions, when I cancel out the negs in the last step, must I also switch the signs of the proceeding variables as well?