Answer
$(1,1),(-1,-1)$
Work Step by Step
$x^2y^2+xy=2\\
2x^2yy'+2y^2x+xy'+y=0\\
2x^2yy'+xy'=-2y^2x-y\\
y'(2x^2y+x)=-2y^2x-y\\
y'=\frac{-2y^2x-y}{2x^2y+x}\\
y'=\frac{-y(2yx+1)}{x(2xy+1)}\\
y'=-\frac{y}{x}$
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.
