Finding the derivative of cos⁡(arcsin⁡x)

With your notation, ${\displaystyle x=\sin u}$, we have
${\displaystyle \frac{d}{dx}\left[\cos \left({\sin }^{-1}x\right)\right]=-\sin \left({\sin }^{-1}x\right)\cdot \frac{d}{dx}\left[{\sin }^{-1}x\right]=-x\frac{d}{dx}\left[{\sin }^{-1}x\right]}$
Since
${\displaystyle \frac{dx}{du}=\cos u}$
we have
${\displaystyle \frac{du}{dx}=\frac{1}{\cos u}=\frac{1}{\sqrt{1-{\sin }^{2}u}}=\frac{1}{\sqrt{1-x^2}}}$
Therefore,
${\displaystyle \frac{d}{dx}\left[\cos \left({\sin }^{-1}x\right)\right]=-\frac{x}{\sqrt{1-x^2}}}$

Let ${\displaystyle y=\cos \left(u\right)=\cos \left(\arcsin x\right)}$. You have correctly worked out that du/dx is
${\displaystyle \frac{1}{\sqrt{1-x^2}}}$
but you need to multiply this result by dy/du. The textbooks

Alternative approach:
The range of $\arcsin \text{ is }[-\frac{\pi }{2},\frac{\pi }{2}]$, so the $\cos [\arcsin (x)]$ will be $\ge 0$
Further, if $\theta =\arcsin (x)$, then $\cos (\theta )$ [which by the above statement will be non-negative] will be $\sqrt{1-x^2}$
Therefore, the question immediately reduces to computing
$\frac{d}{dx}\sqrt{1-x^2}$