The derivative of $$\tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)$$ with respect to $$\tan^{-1}\left(\frac{2x\sqrt{1-x^2}}{1-2x^2}\right)$$ at $$x = \frac{1}{2}$$ is:

Let us denote

$$y(x)=\tan^{-1}\!\left(\dfrac{\sqrt{1+x^{2}}-1}{x}\right), \qquad z(x)=\tan^{-1}\!\left(\dfrac{2x\sqrt{1-x^{2}}}{1-2x^{2}}\right).$$

We have to find $$\dfrac{dy}{dz}$$ at $$x=\dfrac12.$$ Since both $$y$$ and $$z$$ are functions of $$x$$, the chain rule gives

$$\frac{dy}{dz}=\frac{\dfrac{dy}{dx}}{\dfrac{dz}{dx}} =\frac{dy/dx}{dz/dx}.$$

Derivative of $$y$$

Write $$u=\dfrac{\sqrt{1+x^{2}}-1}{x},\quad\text{so}\quad y=\tan^{-1}u.$$

The derivative formula $$\dfrac{d}{dx}\tan^{-1}u=\dfrac{1}{1+u^{2}}\dfrac{du}{dx}$$ yields

$$\frac{dy}{dx}=\frac{1}{1+u^{2}}\frac{du}{dx}.$$

To compute $$\dfrac{du}{dx}$$, note that $$u=\frac{\sqrt{1+x^{2}}-1}{x} =(\sqrt{1+x^{2}}-1)\,x^{-1}.$$

Using the product rule, $$\frac{du}{dx}=\frac{d}{dx}\!\bigl(\sqrt{1+x^{2}}-1\bigr)\,x^{-1} +(\sqrt{1+x^{2}}-1)\,\frac{d}{dx}(x^{-1}).$$

Now $$\frac{d}{dx}\sqrt{1+x^{2}} =\frac{1}{2}(1+x^{2})^{-1/2}\cdot2x =\frac{x}{\sqrt{1+x^{2}}},$$ and $$\frac{d}{dx}(x^{-1})=-\frac1{x^{2}}.$$

Thus

$$\frac{du}{dx} =\frac{x}{\sqrt{1+x^{2}}}\cdot\frac1{x} -(\sqrt{1+x^{2}}-1)\frac1{x^{2}} =\frac1{\sqrt{1+x^{2}}} -\frac{\sqrt{1+x^{2}}-1}{x^{2}}.$$

Derivative of $$z$$

Put $$v=\dfrac{2x\sqrt{1-x^{2}}}{1-2x^{2}},\quad\text{so}\quad z=\tan^{-1}v.$$

Again, the same formula gives

$$\frac{dz}{dx}=\frac{1}{1+v^{2}}\frac{dv}{dx}.$$

Write $$N=2x\sqrt{1-x^{2}},\qquad D=1-2x^{2},\qquad v=\frac{N}{D}.$$

Differentiating $$N$$: $$\frac{dN}{dx}=2\sqrt{1-x^{2}} +2x\cdot\frac{-x}{\sqrt{1-x^{2}}} =2\sqrt{1-x^{2}} -\frac{2x^{2}}{\sqrt{1-x^{2}}} =\frac{2(1-2x^{2})}{\sqrt{1-x^{2}}}.$$

Differentiating $$D$$: $$\frac{dD}{dx}=-4x.$$

Using the quotient rule $$\frac{dv}{dx} =\frac{(dN/dx)\,D-N\,(dD/dx)}{D^{2}} =\frac{\dfrac{2(1-2x^{2})}{\sqrt{1-x^{2}}}(1-2x^{2}) +2x\sqrt{1-x^{2}}\,(4x)} {(1-2x^{2})^{2}}.$$

Combining the terms over the common denominator $$\sqrt{1-x^{2}}$$ gives

$$\frac{dv}{dx} =\frac{2(1-2x^{2})^{2}+8x^{2}(1-x^{2})} {\sqrt{1-x^{2}}\,\bigl(1-2x^{2}\bigr)^{2}}.$$

Evaluating every quantity at }$$x=\dfrac12$$

First compute the needed numbers:

$$x=\frac12,\quad x^{2}=\frac14,\quad 1+x^{2}=\frac54,\quad \sqrt{1+x^{2}}=\frac{\sqrt5}{2},$$

$$1-x^{2}=\frac34,\quad \sqrt{1-x^{2}}=\frac{\sqrt3}{2},\quad 1-2x^{2}=\frac12.$$

For the $$y$$-part: $$u=\frac{\sqrt5/2-1}{1/2}=\sqrt5-2,$$ $$u^{2}=(\sqrt5-2)^{2}=9-4\sqrt5,$$ $$1+u^{2}=10-4\sqrt5,$$ $$\frac{du}{dx} =\frac{2}{\sqrt5}-2\sqrt5+4.$$

For the $$z$$-part: $$N=\frac{\sqrt3}{2},\quad D=\frac12,\quad v=\frac{\sqrt3}{2}\big/\frac12=\sqrt3,$$ $$1+v^{2}=1+3=4.$$

The earlier simplified formula for $$\dfrac{dv}{dx}$$ now gives

$$\frac{dv}{dx} =\frac{2}{\dfrac{\sqrt3}{2}\cdot\left(\dfrac12\right)^{2}} =\frac{2}{\dfrac{\sqrt3}{8}} =\frac{16}{\sqrt3}.$$

Assembling }$$\dfrac{dy}{dz}$$

Because $$\frac{dy}{dx}=\frac{1}{1+u^{2}}\frac{du}{dx},\qquad \frac{dz}{dx}=\frac{1}{1+v^{2}}\frac{dv}{dx},$$ we have $$\frac{dy}{dz} =\frac{\dfrac{du}{dx}}{\dfrac{dv}{dx}} \cdot\frac{1+v^{2}}{1+u^{2}} =\frac{\dfrac{2}{\sqrt5}-2\sqrt5+4}{\dfrac{16}{\sqrt3}} \cdot\frac{4}{10-4\sqrt5} =\frac{\sqrt3}{16}\left(4+\frac{2}{\sqrt5}-2\sqrt5\right) \cdot\frac{4}{10-4\sqrt5}.$$

Simplifying step by step, first factor $$2$$ from the bracket:

$$4+\frac{2}{\sqrt5}-2\sqrt5 =2\!\left(2+\frac1{\sqrt5}-\sqrt5\right) =\frac{2(-2+\sqrt5)}{\sqrt5},$$

so that

$$\frac{dy}{dz} =\frac{\sqrt3}{16}\cdot\frac{2(-2+\sqrt5)}{\sqrt5} \cdot\frac{4}{10-4\sqrt5} =\frac{\sqrt3(-2+\sqrt5)}{2\sqrt5(5-2\sqrt5)}.$$

Notice the common factor $$5-2\sqrt5$$ in the denominator: $$5-2\sqrt5= \frac{10\sqrt5-20}{2\sqrt5}= \frac{10(\sqrt5-2)}{2\sqrt5},$$ so

$$\frac{dy}{dz} =\sqrt3\;\frac{\sqrt5-2}{2\sqrt5}\; \frac{2\sqrt5}{10(\sqrt5-2)} =\sqrt3\;\frac{1}{10} =\frac{\sqrt3}{10}.$$

Hence, the correct answer is Option D.