Can you guess the function

First by simple inspection we can see that $f(1) = 0$ . Now as the function is differentiable $\lim _{ h\rightarrow 0 }{ \frac { f\left( x+h \right) -f\left( x \right) }{ h } } =f^{ ' }\left( x \right)$ we can replace $y$ by $x+h$ and $x$ by $x$ in the above differential equation to get. $\lim _{ h\rightarrow 0 }{ \frac { \frac { { x }^{ x } }{ { (x+h) }^{ (x+h) } } f\left( \frac { { (x+h) }^{ x+h } }{ { x }^{ x } } \right) }{ h } } =f^{ ' }\left( x \right)$ Now $\lim _{ h\rightarrow 0 }{ \frac { { x }^{ x } }{ { (x+h) }^{ (x+h) } } } =1$ . Also using $f(1) = 0$ we get. $\lim _{ h\rightarrow 0 }{ \frac { f\left( \frac { { (x+h) }^{ x+h } }{ { x }^{ x } } \right) -f\left( 1 \right) }{ h } } =f^{ ' }\left( x \right)$ Now note that $\lim _{ h\rightarrow 0 }{ \frac { f\left( \frac { { (x+h) }^{ x+h } }{ { x }^{ x } } \right) -f\left( 1 \right) }{ \frac { { (x+h) }^{ x+h } }{ { x }^{ x } } -1 } } =f^{ ' }\left( 1 \right) =1$ . Substituting it back we get $\lim _{ h\rightarrow 0 }{ \frac { \left( \frac { { (x+h) }^{ x+h } }{ { x }^{ x } } \right) -\left( 1 \right) }{ h } } =f^{ ' }\left( x \right)$ After using L'hopital's rule we get $f^{ ' }(x)=\ln { x } +1$ Solving the differential equation and using at x=1 y=0. we get $f(x)=x\ln{ x }$ So $f(e) = e$ and $f( \frac{ 1 }{ e } ) =\frac{ -1 }{ e }$ .So the answer is $\boxed{ 1 }$