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)#