Let $f(x)$ be differentiable function such that $f\left(\dfrac{x+y}{1-xy}\right)=f(x)+f(y) \forall x$ and $y$. If $\underset { x\rightarrow 0 }{ lt } \dfrac { f\left( x \right) ; }{ x } =\dfrac { 1 }{ 3 }$ then $f(1)$ equals& ;

We have,

$f\left( {\dfrac{{x + y}}{{1 - xy}}} \right) = f\left( x \right) + f\left( y \right)\,\,\,\forall {x^2}y$

$\Rightarrow f\left( x \right) = k{\tan ^{ - 1}}\left( x \right)$

$\mathop {\lim }\limits_{x \to 0} \dfrac{{k{{\tan }^{ - 1}}x}}{x} = \dfrac{1}{3}$

$k = \dfrac{1}{3}$

$f\left( x \right) = \dfrac{1}{3}{\tan ^{ - 1}}x$

$f\left( 1 \right) = \dfrac{1 }{{12}}$

Hence, this is the answer.