In general, $$ (f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))} $$ Let's apply that to arcsine:
$$ \arcsin'(x) = \frac{1}{\sin'(\arcsin(x))} \\\ = \frac{1}{\cos(\arcsin(x))} $$
So then the question is "what's the cosine of an angle whose sine is $x$?" Draw a triangle with vertical leg x and horizontal leg $\sqrt{1 - x^2}$, and hypotenuse $1$. (Check that this satisfies Pythagoras!). The sine of this angle is $x$, so its cosine is $\sqrt{1-x^2}$. So
$$ \arcsin'(x) = \frac{1}{\sin'(\arcsin(x))} \\\ = \frac{1}{\sqrt{1-x^2}}. $$
Now you want the derivative of $4\arcsin(x^3)$, so you need to apply the chain rule:
$$ (4\arcsin(x^3))' = 4\arcsin'(x^3) \cdot (3 x^2)\\\ = 4\frac{1}{\sqrt{1-(x^3)^2}} 3 x^2 \\\ = \frac{12x^2}{\sqrt{1-x^6}}. $$