<s>[INST] Determine $\left(1 - \frac{x_{1}}{y_{1}} \right)\left(1 - \frac{x_{2}}{y_{2}} \right)\left(1 - \frac{x_{3}}{y_{3}} \right)$ $(x_{i},y_{i}), i=1,2,3$ are solutions for
$$ x^{3} - 3xy^{2} = 2010 $$
$$ y^{3} - 3yx^{2} = 2009 $$
What is $\left(1 - \frac{x_{1}}{y_{1}} \right)\left(1 - \frac{x_{2}}{y_{2}} \right)\left(1 - \frac{x_{3}}{y_{3}} \right)$?

Attempt:
My approach is that we should look for $\frac{y-x}{y}=$something.
Notice that
$$ x^{3} - y^{3} + 3(yx^{2} - xy^{2}) = 1  $$
$$ (x-y)(x^{2} + y^{2} + xy) + 3xy(x-y) = 1 $$
$$ (y-x)(x^{2} + y^{2} + 4xy)  = -1$$
$$ \frac{y-x}{y} = -\frac{1}{y(x^{2} + y^{2} + 4xy)} (\text{does not seemt to go anywhere})$$
Anther approach if I sum the equations:
$$ x^{3} + y^{3} - 3(xy^{2} + yx^{2}) = 4019 $$
$$ (x + y)(x^{2} + y^{2} - xy) - 3(xy)(x + y) = 4019 $$
$$ (x+y)(x^{2} + y^{2} - 4xy) = 4019 $$
Multiply both we get
$$ (x^{2}-y^{2})((x^{2}+y^{2})^{2} -16 (xy)^{2}) = 4019 $$
???
 [/INST] It is easy to see none of the $y_i$ vanishes.
Let $u = \frac{x}{y}$ and $u_i = \frac{x_i}{y_i}$. $u_i$ are roots of the equation:
$$x^3 - 3xy^2 - \frac{2010}{2009}(y^3 - 3yx^2) = 0
\iff u^3 - 3u - \frac{2010}{2009}(1 - 3u^2) = 0
$$
This implies
$$(u-u_1)(u-u_2)(u - u_3) = u^3 - 3u - \frac{2010}{2009}(1 - 3u^2)$$
The value we want is the value of this expression at $u = 1$.
$$\prod_{i=1}^3\left(1 - \frac{x_i}{y_i}\right) = \prod_{i=1}^3 (1 - u_i)
= 1 - 3 - \frac{2010}{2009}(1-3) = \frac{2}{2009}$$
</s>