The function $y=\dfrac{\log x}{x^3}$ is strictly increasing function for

Given the function $y = \dfrac{\log x}{x^3}$, we find its derivative with respect to $x$ using the quotient rule.

$\dfrac{dy}{dx} = \dfrac{x^3 \cdot \dfrac{d}{dx}(\log x) - \log x \cdot \dfrac{d}{dx}(x^3)}{(x^3)^2}$

$\dfrac{dy}{dx} = \dfrac{x^3 \cdot \dfrac{1}{x} - \log x \cdot 3x^2}{x^6}$

$\dfrac{dy}{dx} = \dfrac{x^2 - 3x^2 \log x}{x^6}$

$\dfrac{dy}{dx} = \dfrac{x^2(1 - 3 \log x)}{x^6} = \dfrac{1 - 3 \log x}{x^4}$

For the function to be strictly increasing, we require $\dfrac{dy}{dx} > 0$.

Since $x^4 > 0$ for all $x > 0$, the condition simplifies to $1 - 3 \log x > 0$.

$\Rightarrow 1 > 3 \log x$
$\Rightarrow \log x < \dfrac{1}{3}$
$\Rightarrow x < e^{\dfrac{1}{3}}$
Since the domain of $\log x$ is $x > 0$, the function is strictly increasing for $0 < x < e^{\dfrac{1}{3}}$.

Answer: $0 < x < e^{\dfrac{1}{3}}$