Докажите тождество № 5

Решите задачу:

$\frac{ \frac{1}{x^2}+ \frac{1}{y^2} }{(x+y)^2} + \frac{ \frac{2}{x}+ \frac{2}{y} }{(x+y)^3} = \frac{1}{x^2y^2} \\ \\ \frac{ \frac{x^2+y^2}{x^2y^2} }{(x+y)^2} + \frac{ \frac{2x+2y}{xy} }{(x+y)^3} = \frac{1}{x^2y^2} \\ \\ \frac{x^2+y^2 }{x^2y^2(x+y)^2} + \frac{ 2x+2y}{xy(x+y)^3} = \frac{1}{x^2y^2} \\ \\ \frac{(x^2+y^2)(x+y) }{x^2y^2(x+y)^3} + \frac{ (2x+2y)xy}{x^2y^2(x+y)^3} = \frac{1}{x^2y^2} \\ \\ \frac{(x^2+y^2)(x+y)+(2x+2y)xy}{x^2y^2(x+y)^3} = \frac{1}{x^2y^2}$

$\frac{x^3+xy^2+x^2y+y^3+2x^2y+2xy^2} {x^2y^2(x+y)^3}= \frac{1}{x^2y^2} \\ \\\frac{x^3+3xy^2+3x^2y+y^3} {x^2y^2(x+y)^3}= \frac{1}{x^2y^2} \\ \\ \frac{(x+y)^3} {x^2y^2(x+y)^3}= \frac{1}{x^2y^2} \\ \\ \frac{1}{x^2y^2} =\frac{1}{x^2y^2}$