13. 若$(x + 3)^{2}+\vert y - 4\vert = 0$,求$[\frac{1}{2(x - y)}]^{2}\cdot(\frac{2xy^{2}}{x + y})^{3}÷(\frac{xy^{3}}{x^{2}-y^{2}})^{2}$的值.

$-6$

由$(x + 3)^2 + |y - 4| = 0$，得$x + 3 = 0$且$y - 4 = 0$，解得$x = -3$，$y = 4$。
化简原式：
$\begin{aligned}&\left[\frac{1}{2(x - y)}\right]^2 \cdot \left(\frac{2xy^2}{x + y}\right)^3 ÷ \left(\frac{xy^3}{x^2 - y^2}\right)^2\\=&\frac{1}{4(x - y)^2} \cdot \frac{8x^3y^6}{(x + y)^3} ÷ \frac{x^2y^6}{(x + y)^2(x - y)^2}\\=&\frac{1}{4(x - y)^2} \cdot \frac{8x^3y^6}{(x + y)^3} \cdot \frac{(x + y)^2(x - y)^2}{x^2y^6}\\=&\frac{8x^3y^6(x + y)^2(x - y)^2}{4(x - y)^2(x + y)^3x^2y^6}\\=&\frac{2x}{x + y}\end{aligned}$
将$x = -3$，$y = 4$代入$\frac{2x}{x + y}$，得$\frac{2×(-3)}{-3 + 4} = \frac{-6}{1} = -6$。

14. (1)计算：$(-\frac{2a}{b^{2}})^{2}\cdot(\frac{2b}{a})^{2}÷(-\frac{2b^{2}}{a})$；
(2)已知$x + 2y = 4$,$x - 2y = -1$,先化简,再求值：$\frac{x^{2}-4y^{2}}{x^{2}+2xy + y^{2}}÷(\frac{x + 2y}{x + y})^{2}$；
(3)已知$x - y = \frac{1}{2}$,$xy = -\frac{3}{4}$,先分解因式,再求值：$x^{3}y - 2x^{2}y^{2}+xy^{3}$.

(1)
原式$=\frac{4a^{2}}{b^{4}}\cdot\frac{4b^{2}}{a^{2}}÷(-\frac{2b^{2}}{a})$
$=\frac{16}{b^{2}}\cdot(-\frac{a}{2b^{2}})$
$=-\frac{8a}{b^{4}}$
(2)
$\because x + 2y = 4$，$x - 2y = -1$，
原式$=\frac{(x + 2y)(x - 2y)}{(x + y)^{2}}÷\frac{(x + 2y)^{2}}{(x + y)^{2}}$
$=\frac{(x + 2y)(x - 2y)}{(x + y)^{2}}\cdot\frac{(x + y)^{2}}{(x + 2y)^{2}}$
$=\frac{x - 2y}{x + 2y}$
把$x + 2y = 4$，$x - 2y = -1$代入得：原式$=-\frac{1}{4}$
(3)
$\because x - y = \frac{1}{2}$，$xy = -\frac{3}{4}$，
原式$=xy(x^{2}-2xy + y^{2})$
$=xy(x - y)^{2}$
把$x - y = \frac{1}{2}$，$xy = -\frac{3}{4}$代入得：
原式$=-\frac{3}{4}×(\frac{1}{2})^{2}=-\frac{3}{16}$