$x^3 + y^3 +3x^2 y^2 =x^3y^3$ 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.

The following is equivalent to @NN2's solution, only written a bit differently.
First substitute $x=1/a, y=1/b$, so that the equation becomes
$$ \tag{*}
 a^3+b^3+3ab - 1 = 0 \, .
$$
We are looking for the possible values of $a+b$, this suggests to introduce $S=a+b$ and $P = ab$. Then
$$
 0 = a^3+b^3+3ab - 1 = S^3+3PS + 3P -1 = (S-1)(S^2 + S-3P+1) \, .
$$
So $S=1$ is one possible value for $a+b$. Conversely, all $(a, b)$ with $a+b=1$ satisfy the equation $(*)$.
In order to determine the possible zeros of the second factor we use that $P \le S^2/4$ from the inequality between geometric and arithmethic mean:
$$
0 = S^2 + S-3P+1 \ge S^2 + S-\frac 34 S^2+1 = \frac 14 (S+2)^2 \ge 0
$$
is only possible if $S=-2$ and $P=1$, i.e. if $(a, b) = (-1, -1)$.
Therefore $S=a+b=1/x+1/y$ can only take the values $1$ and $-2$, and both values do actually occur.
