#### Answer

$\dfrac{-3}{y+3}$

#### Work Step by Step

Simplify by multiplying the LCD (which is $y(y+3)$) to both the numerator and the denominator to obtain:
$\require{cancel}
=\dfrac{y(y+3)\left(\frac{1}{y+3}-\frac{1}{y}\right)}{y(y+3)(\frac{1}{x})}
\\=\dfrac{y(y+3)(\frac{1}{y+3})-y(y+3)(\frac{1}{y})}{y(y+3)(\frac{1}{y})}
\\=\dfrac{y\cancel{(y+3)}(\frac{1}{\cancel{y+3}})-\cancel{y}(y+3)(\frac{1}{\cancel{y}})}{\cancel{y}(y+3)(\frac{1}{\cancel{y}})}
\\=\dfrac{y-(y+3)}{y+3}
\\=\dfrac{y-y-3}{y+3}
\\=\dfrac{-3}{y+3}$