<s>[INST] $x^3 + y^3 +3x^2 y^2 =x^3y^3$ 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.
 [/INST] $$x^3 + y^3 + 3 x^2 y^2 = x^3 y^3\tag{1}$$
$$\frac{x+y}{xy}\tag{2}$$
$$x^3 + y^3 + 3 x^2 y^2 - x^3 y^3=0$$
can be factorized$^1$ as
$$(x+y-xy) \left(x^2 y^2+x^2 y+x^2+x y^2-x y+y^2\right)=0$$
$x^2 y^2+x^2 y+x^2+x y^2-x y+y^2=0$ has solution$^2$ given by $x=y=-1$ we have $$\frac{x+y}{xy}=-2$$, the solutions of $(1)$ are only those of $x+y-xy=0$ that is $x+y=xy$ and finally
$$\frac{x+y}{xy}=1$$

$(^1)$
$$x^3 + y^3 + 3 x^2 y^2 - x^3 y^3=(x+y)^3-3x^2y-3xy^2+ 3 x^2 y^2 - x^3 y^3=$$
$$=\left[(x+y)^3- x^3 y^3\right]+\left[-3x^2y-3xy^2+ 3 x^2 y^2 \right]=$$
$$=(x+y-xy)\left[(x+y)^2+xy(x+y)+x^2y^2)\right]-3 x y (x+y-x y)=$$
$$=(x+y-xy)\left[(x+y)^2+xy(x+y)+x^2y^2)-3xy\right]=(x+y-xy)\left(x^2 y^2+x^2 y+x^2+x y^2-x y+y^2\right)$$
$(^2)$
$$x^2 y^2+x^2 y+x^2+x y^2-x y+y^2=0$$
set $y=kx$. We get
$$k^2 x^4+k^2 x^3+k^2 x^2+k x^3-k x^2+x^2=0\to x^2 \left(k^2 x^2+k^2 x+k x+k^2-k+1\right)=0$$
As $x\ne 0$ we have
$$k^2 x^2+k^2 x+k x+k^2-k+1=0$$
the discriminant of this equation is $D=-k^4+2 k^3-k^2=-k^2(k-1)^2$
$D\ge 0\to k=1 $ (solution $k=0$ is discarded)
if $k=1$ we have $x^2+2x+1=0\to x=-1$ thus the solutions are
$$x=-1;\;y=-1$$
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