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