Let $f$ be a differentiable function defined for all $x\in \mathbb{R}$ such that $f(x^3)=x^5$ for all $x\ne 0$. Then the value of $\frac{df}{dx}(8)$ is:

Given $f(x^3)=x^5$.Differentiate both sides with respect to $x$:
$$\frac{d}{dx}f(x^3)=\frac{d}{dx}{x}^5$$
$$f^{\prime }(x^3)\cdot 3x^2=5x^4$$
$$f^{\prime }(x^3)=\frac{5x^4}{3x^2}=\frac{5}{3}{x}^2$$
We need to find $\frac{df}{dx}(8)$, which is $f^{\prime }(8)$.To get $8$ inside the function, set $x^3=8⟹x=2$.Substitute $x=2$ into the derived formula:
$$f^{\prime }(8)=\frac{5}{3}(2{)}^2=\frac{5}{3}\cdot 4=\frac{20}{3}$$