Z = log($e^x+ e^y$)

(1) $\frac{∂z}{∂x}$ = $\frac{e^x}{((e^x+ e^y))}$ $\frac{(∂^2 Z)}{(∂x^2)}$ =$\frac{(e^x(e^x+e^y)-e^x(e^x))}{(e^x+e^y)^2}$ = $\frac{(e^2x+e^xy-e^2x)}{(e^x+ e^y)^2}$

r= $\frac{(∂^2 Z)}{(∂x^2)}$ = $\frac{e^xy}{(e^x+ e^y )^2}$ ………….(1)

(2) $\frac{∂z}{∂y}$ = $\frac{e^y}{((e^x+ e^y))}$ $\frac{(∂^2 Z)}{(∂y^2)}$ = $\frac{(e^y(e^x+ e^y)-e^y(e^y ))}{(e^x+e^y)^2}$ = $\frac{(e^2y+e^xy-e^2y)}{(e^x+ e^y)^2}$

t=$\frac{(∂^2 Z)}{(∂y^2)}$ = $\frac{e^xy}{(e^x+ e^y )^2}$ ………….(2)

(3) $\frac{∂z}{∂x}$ = $\frac{e^x}{((e^x+ e^y))}$ ,s =$\frac{(∂^2 Z)}{(∂x ∂y)}$ = $\frac{e^xy}{(e^x+ e^y)^2}$ ……………..(3)

From (1) , (2) and (3) we get ,

rt = $\frac{e^xy}{(e^x+e^y)^2}$ × $\frac{e^xy}{(e^x+ e^y)^2}$ = $(\frac{e^xy}{(e^x+ e^y)^2)})^2$ = $\frac{e^2xy}{(e^x+e^y)^2}$ …………..(4)

$s^2$ = $(\frac{e^xy}{(e^x+ e^y)^2})^2$ = $\frac{e^2xy}{(e^x+ e^y)^2}$ ……………….(5)

From (4) and(5) we get,

rt-$s^2$ = $\frac{e^2xy}{(e^x+e^y)^2}$ - $\frac{e^2xy}{(e^x+ e^y)^2} = 0. Hence proved rt-$s^2$ = 0