## Calculus: Early Transcendentals 8th Edition

$(1,1),(-1,-1)$
$x^2y^2+xy=2\\ 2x^2yy'+2y^2x+xy'+y=0\\ 2x^2yy'+xy'=-2y^2x-y\\ y'(2x^2y+x)=-2y^2x-y\\ y'=\frac{-2y^2x-y}{2x^2y+x}\\ y'=\frac{-y(2yx+1)}{x(2xy+1)}\\ y'=-\frac{y}{x}$ In order to find the points where the gradient is $-1$, we equal $-1$ to $y'$ $-1=-\frac{y}{x}\\ y=x$ Then we equate $x=y$ to the curve $y^2y^2+yy=2\\ y^4+y^2=2\\ y^2(y^2+1)=2\\ y=1\Longrightarrow x=1\\ y=-1\Longrightarrow x=-1$ Therefore, $(1,1); (-1,-1)$ are the points where the gradient of the tangents is -1.