Given that, $x$ is a real number.
Now, $\dfrac{x}{\sqrt{1+x^{4}}} = \dfrac{1}{\frac{\sqrt{1+x^{4}}}{x}} = \dfrac{1}{\sqrt{\frac{{1+x^{4}}}{x^{2}}}} = \dfrac{1}{\sqrt{\frac{1}{x^{2}} + x^{2}}} \quad \longrightarrow (1) $
We know that, $ \boxed{\text{AM} \geq \text{GM}}$
$ \Rightarrow \dfrac{x^{2} + \dfrac{1}{x^{2}}}{2} \geq \sqrt{x^{2} \cdot \dfrac{1}{x^{2}}} $
$ \Rightarrow \boxed {x^{2} + \frac{1}{x^{2}} \geq 2} $
From equation $(1),$ we get
$ \dfrac{x}{\sqrt{1+x^{4}}} = \dfrac{1}{\sqrt{\underbrace{x^{2}+\frac{1}{x^{2}}}_{2}}}$
$ = \dfrac{1}{\sqrt{2}} \quad [ \because \text{For the maximum value}] $
$\therefore$ The maximum possible value of $\dfrac{x}{\sqrt{1+x^{4}}}$ is $\dfrac{1}{\sqrt{2}}.$
Correct Answer$: \text{C}$