If $x+y=3$ and $\frac{1}{x}+\frac{1}{y}=-\frac{3}{10}$, then the value of $\left(x^2+y^2\right)$ is :

29

$x+y=3$

$\frac{1}{x}+\frac{1}{y}=-\frac{3}{10}$

\(\frac{y + x}{xy}\) = -\(\frac{3}{10}\) 

\(\frac{3}{xy}\) = -\(\frac{3}{10}\) 

xy = -10

( a + b )² = a² + b² + 2ab

$\left(x^2+y^2\right)$ =  ( 3 )² - 2 × -10

$\left(x^2+y^2\right)$ = 29