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

Because $x,y \ne 0$, from $x^3 + y^3 +3x^2 y^2 =x^3y^3$, we have
$$\frac{1}{x^3}+\frac{1}{y^3} +\frac{3}{xy} =1 \iff\frac{1}{x^3}+\frac{1}{y^3}  =1-\frac{3}{xy} \tag{1}$$
But we have also
$$(\frac{1}{x}+\frac{1}{y})^3 = (\frac{1}{x^3}+\frac{1}{y^3}) + \frac{3}{xy}(\frac{1}{x}+\frac{1}{y}) \tag{2}$$
From (1) and (2), we deduce
\begin{align}
&(\frac{1}{x}+\frac{1}{y})^3 = (1-\frac{3}{xy}) + \frac{3}{xy}(\frac{1}{x}+\frac{1}{y}) \\
&\iff (\frac{1}{x}+\frac{1}{y})^3- 1=\frac{3}{xy}\left( \frac{1}{x}+\frac{1}{y} - 1  \right) \\
&\iff \left( \frac{1}{x}+\frac{1}{y} - 1  \right)\left( (\frac{1}{x}+\frac{1}{y})^2 +(\frac{1}{x}+\frac{1}{y})+1   \right)=\frac{3}{xy}\left( \frac{1}{x}+\frac{1}{y} - 1  \right) \\
&\iff \left( \frac{1}{x}+\frac{1}{y} - 1  \right)\left( (\frac{1}{x}+\frac{1}{y})^2 +(\frac{1}{x}+\frac{1}{y})+1  -\frac{3}{xy} \right)=0\\
&\iff \left( \frac{1}{x}+\frac{1}{y} - 1  \right)\left( (\frac{1}{x}+\frac{1}{y})^2 -\frac{4}{xy} +(\frac{1}{x}+\frac{1}{y})+1 +\frac{1}{xy}   \right)=0\\
&\iff \left( \frac{1}{x}+\frac{1}{y} - 1  \right)\left( (\frac{1}{x}-\frac{1}{y})^2  +\left( 1+\frac{1}{x}   \right)\left( 1+\frac{1}{y} \right)  \right)  =0\\
&\iff \left( \frac{1}{x}+\frac{1}{y} - 1  \right)\left( ((1+\frac{1}{x})-(1+\frac{1}{y}))^2  +\left( 1+\frac{1}{x}   \right)\left( 1+\frac{1}{y} \right)  \right)  =0\\
&\iff \left( \frac{1}{x}+\frac{1}{y} - 1  \right)\left( \left(1+\frac{1}{x} \right)^2+\left(1+\frac{1}{y} \right)^2  -\left( 1+\frac{1}{x}   \right)\left( 1+\frac{1}{y} \right)  \right)  =0\\
\end{align}
The last equation happens if and only if $\frac{1}{x}+\frac{1}{y} =1$  or $\left( 1+\frac{1}{x}, 1+\frac{1}{y} \right) = (0,0) $. Hence, the two possible values of $\frac{1}{x}+\frac{1}{y}$ is $1$ and $-2$.
