The slope of the tangent line is given by #dy/dx#, which we can calculate by implicit differentiation:
#d/dx(x^2y^2+xy) = 0#
#2x^2ydy/dx+2xy^2 +y +xdy/dx=0#
#(x^2y+x)dy/dx= -xy^2 -y#
Since from the original equation we can see that #xy!=0#, divide both sides by #xy#
#(x+1/y)dy/dx = -(y+1/x)#
#(xy+1)/y dy/dx =- (xy+1)/x#
We can see that #xy != -1#, otherwise, #(xy)^2+xy# would be null, in contradiction with the original equation, so:
#dy/dx =-y/x#
And we have:
#-y/x = -1#
which means:
#x=y#
Substituting this in the original equation we find the values of #x# for which #dy/dx = -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)#
