The derivative of $\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ w.r.t. $\tan ^{-1}\left(\frac{2 x \sqrt{1-x^2}}{1-2 x^2}\right)$ at $x=0$ is

Let $u = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$ and $v = \tan^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)$.
To find $\frac{du}{dv}$ at $x=0$, first simplify both expressions using trigonometric substitutions.

Substitute $x = \tan\theta$ for $u$:
$u = \tan^{-1}\left(\frac{\sqrt{1+\tan^2\theta}-1}{\tan\theta}\right) = \tan^{-1}\left(\frac{\sec\theta-1}{\tan\theta}\right) = \tan^{-1}\left(\frac{1-\cos\theta}{\sin\theta}\right)$.

Using half-angle identities $1-\cos\theta = 2\sin^2(\theta/2)$ and $\sin\theta = 2\sin(\theta/2)\cos(\theta/2)$:
$u = \tan^{-1}\left(\frac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)}\right) = \tan^{-1}\left(\tan(\theta/2)\right) = \frac{\theta}{2}$.

Since $\theta = \tan^{-1}x$, we have $u = \frac{1}{2}\tan^{-1}x$, and $\frac{du}{dx} = \frac{1}{2} \cdot \frac{1}{1+x^2}$.

Now substitute $x = \sin\phi$ for $v$:
$v = \tan^{-1}\left(\frac{2\sin\phi\sqrt{1-\sin^2\phi}}{1-2\sin^2\phi}\right) = \tan^{-1}\left(\frac{2\sin\phi\cos\phi}{\cos2\phi}\right) = \tan^{-1}\left(\frac{\sin2\phi}{\cos2\phi}\right) = \tan^{-1}(\tan2\phi) = 2\phi$.

Since $\phi = \sin^{-1}x$, we have $v = 2\sin^{-1}x$, and $\frac{dv}{dx} = \frac{2}{\sqrt{1-x^2}}$.

By the chain rule, $\frac{du}{dv} = \frac{du/dx}{dv/dx} = \frac{1/(2(1+x^2))}{2/\sqrt{1-x^2}} = \frac{\sqrt{1-x^2}}{4(1+x^2)}$.

At $x=0$:
$\left.\frac{du}{dv}\right|_{x=0} = \frac{\sqrt{1-0^2}}{4(1+0^2)} = \frac{1}{4}$.

The derivative $\frac{du}{dv}$ evaluated at $x=0$ is $\frac{1}{4}$.