\[\frac{x + 4y}{y} = 12\]
\[\frac{x}{y} + \frac{4y}{y} = 12\]
\[\frac{x}{y} + 4 = 12\]
\[\frac{x}{y} = 8\]
\[x = 8y.\]
\[\frac{x^{2} + 16y^{2}}{x^{2} - 3xy} =\]
\[= \frac{(8y)^{2} + 16y^{2}}{(8y)^{2} - 3y \cdot 8y} =\]
\[= \frac{64y^{2} + 16y^{2}}{64y^{2} - 24y^{2}} = \frac{80y^{2}}{40y^{2}} = 2.\]
\[Ответ:2.\]