The slope of the tangent line is given by dy/dxdydx, which we can calculate by implicit differentiation:
d/dx(x^2y^2+xy) = 0ddx(x2y2+xy)=0
2x^2ydy/dx+2xy^2 +y +xdy/dx=02x2ydydx+2xy2+y+xdydx=0
(x^2y+x)dy/dx= -xy^2 -y(x2y+x)dydx=−xy2−y
Since from the original equation we can see that xy!=0xy≠0, divide both sides by xyxy
(x+1/y)dy/dx = -(y+1/x)(x+1y)dydx=−(y+1x)
(xy+1)/y dy/dx =- (xy+1)/xxy+1ydydx=−xy+1x
We can see that xy != -1xy≠−1, otherwise, (xy)^2+xy(xy)2+xy would be null, in contradiction with the original equation, so:
dy/dx =-y/xdydx=−yx
And we have:
-y/x = -1−yx=−1
which means:
x=yx=y
Substituting this in the original equation we find the values of xx for which dy/dx = -1dydx=−1
{(x^2y^2+xy=2),(x=y):}
x^4+x^2 = 2
x^4+x^2-2=0
Solving this as a second degree equation with unknown x^2:
x^2 = (-1+-sqrt(1+8))/2
And choosing only the positive solution since x^2 cannot be negative for x in RR:
x^2 = 1
x=+-1
Thus we have the points:
P_1 = (-1,-1) and P_2 = (1,1)