Answer
$r_{xx}=\frac{-1}{(x+y)^2}$
$r_{yy}=\frac{-1}{(x+y)^2}$
$r_{xy}=r_{yx}=\frac{-1}{(x+y)^2}$
Work Step by Step
Take the first partial derivatives of the given function. When taking partial derivative with respect to x, treat y as a constant, and vice versa:
$r_x=\frac{1}{x+y}$
$r_y=\frac{1}{x+y}$
Then take the derivative of the first order partial derivatives to find second partial derivatives:
$r_{xx}=\frac{(x+y)(0)-1}{(x+y)^2}=\frac{-1}{(x+y)^2}$
$r_{yy}=\frac{(x+y)(0)-1}{(x+y)^2}=\frac{-1}{(x+y)^2}$
Second partial derivatives of first order partial derivative of x with respect to y and y with respect to x are the same:
$r_{xy}=r_{yx}=\frac{(x+y)(0)-1}{(x+y)^2}=\frac{-1}{(x+y)^2}$
