1. Factorización de la primera ecuación:
La ecuación $4x^2 - 3xy - y^2 = 0$ es homogénea. Factorizamos:
$$ (4x + y)(x - y) = 0 $$
Esto nos da dos casos para sustituir en la segunda ecuación.

2. Análisis por casos:

Caso 1: $y = x$
Sustituimos en $32x^2 - 36xy + 9y^2 = 6$:
$$ 32x^2 - 36(x)(x) + 9(x)^2 = 6 \implies (32 - 36 + 9)x^2 = 6 \implies 5x^2 = 6 $$
$$ x^2 = \frac{6}{5} \implies x = \pm \sqrt{\frac{6}{5}} = \pm \frac{\sqrt{30}}{5} $$
Como $y = x$, entonces $y = \pm \frac{\sqrt{30}}{5}$.

Caso 2: $y = -4x$
Sustituimos en la segunda ecuación:
$$ 32x^2 - 36x(-4x) + 9(-4x)^2 = 6 \implies 32x^2 + 144x^2 + 144x^2 = 6 $$
$$ 320x^2 = 6 \implies x^2 = \frac{6}{320} = \frac{3}{160} $$
$$ x = \pm \sqrt{\frac{3}{160}} = \pm \frac{\sqrt{30}}{40} $$
Para $y = -4x$:
$$ y = -4 \left( \pm \frac{\sqrt{30}}{40} \right) = \mp \frac{\sqrt{30}}{10} $$

Representación de soluciones:
$$ \begin{array}{ll} (x_1, y_1) = \left( \frac{\sqrt{30}}{5}, \frac{\sqrt{30}}{5} \right) & (x_2, y_2) = \left( -\frac{\sqrt{30}}{5}, -\frac{\sqrt{30}}{5} \right) \\ (x_3, y_3) = \left( \frac{\sqrt{30}}{40}, -\frac{\sqrt{30}}{10} \right) & (x_4, y_4) = \left( -\frac{\sqrt{30}}{40}, \frac{\sqrt{30}}{10} \right) \end{array} $$

$$ \boxed{(x, y) \in \left\{ \pm \left( \frac{\sqrt{30}}{5}, \frac{\sqrt{30}}{5} \right), \pm \left( \frac{\sqrt{30}}{40}, -\frac{\sqrt{30}}{10} \right) \right\}} $$