Ответ:

[tex]\displaystyle -\frac{y}{x-y}[/tex]

Объяснение:

[tex]\displaystyle \left(\frac{y^2}{x^3-xy^2}+\frac{1}{x+y}\right) : \left(\frac{x-y}{x^2+xy}-\frac{x}{xy+y^2} \right)=\\\\\\\left(\frac{y^2}{x(x^2-y^2)}+\frac{1}{x+y}\right) : \left(\frac{x-y}{x(x+y)}-\frac{x}{y(x+y)} \right)=[/tex]

[tex]\displaystyle\left(\frac{y^2}{x(x-y)(x+y)}+\frac{1}{x+y}\right) : \left(\frac{y(x-y)}{xy(x+y)}-\frac{x^2}{xy(x+y)} \right)=\\\\\\\left(\frac{y^2}{x(x-y)(x+y)}+\frac{x(x-y)}{x(x-y)((x+y)}\right) : \frac{y(x-y)-x^2}{xy(x+y)}=\\\\\\\frac{y^2+x(x-y)}{x(x-y)(x+y)}: \frac{xy-y^2-x^2}{xy(x+y)}=\\\\\\\frac{y^2+x^2-xy}{x(x-y)(x+y)}\cdot \frac{xy(x+y)}{xy-y^2-x^2}=\\\\\\\frac{y^2+x^2-xy}{x-y}\cdot \frac{y}{-(xy+y^2+x^2)}=-\frac{y}{x-y}[/tex]