For $x \in (0, 3/2)$, let $f(x) = \sqrt{x}$, $g(x) = \tan x$ and $h(x) = \dfrac{1 - x^2}{1 + x^2}$. If $\phi(x) = (h \circ f \circ g)(x)$, then $\phi\left(\dfrac{\pi}{3}\right)$ is equal to :