Answer
$r_{xx}=\dfrac{-1}{(x+y)^2} ; \\r_{yy}=\dfrac{-1}{(x+y)^2}$
and
$r_{xy}=r_{yx}=\dfrac{-1}{(x+y)^2}$
Work Step by Step
Our aim is to take the first partial derivative of the given function $r(x,y)$ with respect to $x$, by treating $y$ as a constant and vice versa:
$r_x=\dfrac{1}{x+y} $
$r_y=\dfrac{1}{x+y}$
Now, compute second partial derivatives.
$r_{xx}=\dfrac{(x+y)(0)-1}{(x+y)^2}=\dfrac{-1}{(x+y)^2} ; \\r_{yy}=\dfrac{(x+y)(0)-1}{(x+y)^2}=\dfrac{-1}{(x+y)^2}$
and
$r_{xy}=r_{yx}=\dfrac{(x+y)(0)-1}{(x+y)^2}=\dfrac{-1}{(x+y)^2}$
