Given

$\rm \frac{a^{2} + b^{2}}{ab} - \frac{b^{2}}{a(a+b)} - \frac{a^{2}}{b (a+b)}$

Let us identify the terms as first, second, and third from the left to the right. Then, multiply and divide the first term by \( \rm (a+b) \), the second term by \( \rm b \), and the third term by \( \rm a \).

$\rm = \frac{a^{2} + b^{2}}{ab} \cdot \frac{a +b}{a+b} -  \frac{b^{2}}{a(a+b)} \cdot \frac{b}{b} - \frac{a^{2}}{b (a+b)} \cdot \frac{a}{a}$

$\rm =  \frac{(a^{2} + b^{2})(a +b)}{ab(a+b)}  - \frac{b^{3}}{ab(a+b)} - \frac{a^{3}}{b(a+b)}$

The denominators of the first, second, and third terms are common. So, we simplify the expression as shown below.

$\rm = \frac{ (a+b)(a^{2} + b^{2}) - b^{3} - a^{3}}{ab (a +b)}$

$\rm = \frac{a (a^{2} + b^{2}) + b(a^{2} + b^{2}) - b^{3} -  a^{3}}{ab ( a + b)}$

$\rm = \frac{a^{3} + ab^{2} + a^{2}b + b^{3} - b^{3} - a^{3}}{ab ( a + b)}$

$\rm = \frac{ ab^{2} + a^{2} b}{ab ( a + b)}$

$\rm = \frac{ ab (a + b) }{ab ( a + b)}$

$\rm = 1$

Hence, $$\rm \frac{a^{2} + b^{2}}{ab} - \frac{b^{2}}{a(a+b)} - \frac{a^{2}}{b (a+b)} = 1$$