Derivative of an inverse function

1. $f'(x)=3x^2+1>0$, so $f$ is strictly increasing, and $f(x)\to\pm\infty$ at $\pm\infty$, so $f$ is a continuous bijection onto $\mathbb{R}$. We cannot easily write $f^{-1}$ explicitly, but we can find where it takes the value $2$: $f(1)=2$, so $f^{-1}(2)=1$. Then
$$(f^{-1})'(2)=\frac1{f'(1)}=\frac14.$$

2. Let $x=\arcsin y\in(-\frac\pi2,\frac\pi2)$, so $y=\sin x$ and $\cos x>0$. Then
$$\arcsin'(y)=\frac1{\cos x}=\frac1{\sqrt{1-\sin^2x}}=\frac1{\sqrt{1-y^2}}.$$

Remark. The geometric meaning: the graph of $f^{-1}$ is the mirror image of that of $f$ across the line $y=x$, so slopes are inverted.