$x^3 + y^3 +3x^2 y^2 =x^3y^3$ Find all possible values to $\frac{x+y}{xy}$ $x,y \in \mathbb{R}\setminus\{0\}$
Indeed the original question said :Find all possible values to: $$\frac{1}{x} + \frac{1}{y}$$
But it’s the same thing.
My Attempt:
$$x^3 + y^3 +3x^2 y^2 =x^3y^3 \iff (x+y)(x^2-xy+y^2)=x^3y^3 -3x^2y^2$$
$$\frac{x+y}{xy}=\frac{x^2y^2-3xy}{x^2-xy+y^2}$$
But i don’t know what to do now.
Please if you know the key to this type of problems ‘Find all possible values’ post it.

Hint:
Let $\dfrac{x+y}{xy}=k$
$$(xy)^3=3(xy)^2+(x+y)^3-3xy(x+y)$$
$$\iff(xy)^3(1-k^3)-3(xy)^2+3xy(kxy)=0$$
$$\iff(xy)^3(1-k^3)+3(xy)^2(k-1)=0$$
As $xy\ne0$ $$(xy)(1-k^3)=3(1-k)$$
Clearly, $k=1$ is a solution.
Otherwise,  $$xy=\dfrac3{1+k+k^2}$$
