What are the possible real values of $\frac{1}{x} + \frac{1}{y}$ given $x^3 +y^3 +3x^2y^2 = x^3y^3$? Let $x^3 +y^3 +3x^2y^2 = x^3y^3$ for $x$ and $y$ real numbers different from $0$.
Then determine all possible values of $\frac{1}{x} + \frac{1}{y}$
I tried to factor this polynomial but there's no a clear factors

The equation is equivalent to
$$x^3+y^3+(-xy)^3 -3xy(-xy)=0 \tag{1}$$
Applying this formula
$$a^3+b^3+c^3 -3abc= \frac{1}{2}(a+b+c)((a-b)^2+(b-c)^2+(c-a)^2)$$
for $(a,b,c) = (x,y,-xy)$
then $(1)$ holds true if and only if either of these 2 equations holds true $$x+y-xy = 0\tag{2}$$ $$x=y=-xy \tag{3}$$
If $(2)$ holds true then $\frac{1}{x}+\frac{1}{y} = 1$
if $(3)$ holds true then $x = y = -1 \implies \frac{1}{x}+\frac{1}{y} = -2$
Conclusion:
$$\frac{1}{x}+\frac{1}{y} = 1 \text{ or } \frac{1}{x}+\frac{1}{y} =-2$$
