If for all $x, y$ the function f is defined by; $f(x)+f(y)+f(x)\cdot f(y)=1$ and $f(x) > 0$.When $f(x)$ is differentiable $f'(x)=$,

Here, $f(x)+f(y)+f(x)\cdot f(y)=1$ .....(i)
Substitute $x=y=0$, we get

$2f(0)+\left \{f(0)\right \}^2=1\Rightarrow \left \{f(0)\right \}^2+2f(0)-1=0$

$f(0)=\frac {-2\pm \sqrt {4+4}}{2}=-1-\sqrt 2$ and $-1+\sqrt 2$

As $f(0) > 0\Rightarrow f(0)=\sqrt 2-1$ [neglecting $f(0)=-1-\sqrt 2$ as f(0) is positive]

Again, putting $y=x$ is Eq, (i), $2f(x)+\left \{f(x)\right \}^2=1$

On differentiating w.r.t. x, $2f'(x)+2f(x)f'(x)=0$
$2f'(x)\left \{1+f(x)\right \}=0\Rightarrow f'(x)=0$ because $f(x) > 0$
Thus, $f'(x)=0$ when $f(x) > 0$