$z=\log (x^2+y^2) + \dfrac {x^2 + y^2 }{ x+y} - 2 \log (x+y) \ $ $\dfrac {\partial z} { \partial x} = \dfrac {1}{x^2 + y^2 } \dfrac {\partial (x^2 + y^2 )}{\partial x} + \dfrac {2x ( x+y)- (x^2 + y^2 )}{(x+y)^2} - \dfrac {2}{x+y} \dfrac {\partial (x+y)}{\partial x}$ $= \dfrac {2x}{x^2 + y^2} + \dfrac {2x^2 + 2xy - x^2 - y^2}{(x+y)^2 } - \dfrac {2}{x+y}= \dfrac {2x}{x^2 + y^ 2} + \dfrac {x^2 + 2xy - y^2}{(x+y)^2} - \dfrac {2}{x+y} \ $ $\therefore x \dfrac {\partial z} {\partial x} = \dfrac {2x^3} {x^2+y^2} + \dfrac {x^3 + 2x^2 - xy^2} { (x+y)^2} - \dfrac {2x}{x+y} $ $Similarly , \ y \dfrac {\partial z}{\partial y} = \dfrac {2y^2}{x^2 + y^2} + \dfrac {y^3 + 2y^2x-x^2y} { (x+y)^2} - \dfrac {2y}{x+y} $ $\therefore x \dfrac {\partial z}{\partial x} + y \dfrac {\partial z}{\partial y} $ $= \dfrac {2x^2} { x^2 + y^2} + \dfrac {x^3 + 2x^2 - xy^2} { (x+y)^2} - \dfrac {2x}{x+y} + \dfrac {2y^2}{x^2+y^2} + \dfrac {y^3 + 2y^2 x - x^2 y} { (x+y)^2} - \dfrac {2y}{x+y}$ $= \dfrac {2 (x^2 + y^2)}{x^2 + y^2 }+ \dfrac {x^3 + x^2 y + xy^2 + y^3}{(x+y)^2}- \dfrac {2 (x+y)}{x+y} $ $= 2 + \dfrac {x^2 (x+y)+ y^2 (x+y)}{(x+y)^2} - 2 $ $= \dfrac {(x+y)(x^2 + y^2)} { (x+y)^2}$ $= \dfrac {x^2 + y^2}{x+y}$