Finding integer roots of an equation 
Find all negative integer solutions of x and y such that
$$
x^3 + 3y^2 = xy^2 + 27
$$

To begin, I isolated $x$ and $y$ to see if there is anything I could do and got
$$
y = \pm \sqrt{x^2+3x+9} \\
x = \dfrac{1}{2} \left(-3 \pm \sqrt{y^2-27}\right)
$$
Since $x$ and $y$ are negative here I thought I could remove the positive branch and got
$$
y = - \sqrt{x^2+3x+9} \\
x = - \dfrac{1}{2} \left(3 + \sqrt{y^2-27}\right)
$$
And I really don't know what to do at this point. I don't have much knowledge on finding integer solutions, any help is appreciated.

The equation $x^3 + 3y^2 = xy^2 + 27$ can be written as
$$x^3-27+3y^2-xy^2=0$$
$$(x-3)(x^2+3x+9)+y^2(3-x)=0$$
$$(x-3)(x^2+3x+9-y^2)=0$$
Now, $x^2+3x+9-y^2=0$ can be written as
$$\bigg(x+\frac 32\bigg)^2+\frac{27}{4}-y^2=0$$
Multiplying the both sides by $4$ gives
$$(2x+3)^2+27-(2y)^2=0$$
and so
$$(2y-2x-3)(2y+2x+3)=27$$
I think that you can continue from here.
