This problem is not solved.
$$ \begin{align} f(x) &=\log\ \sqrt{\frac{1+\sqrt{2}x +x^2}{1-\sqrt{2}x +x^2}}+\tan^{-1}\left(\frac{\sqrt{2}x}{1-x^2}\right) \cr \frac{df}{dx}&=\mathord? \end{align} $$
This problem is not solved.
$$ \begin{align} f(x) &=\log\ \sqrt{\frac{1+\sqrt{2}x +x^2}{1-\sqrt{2}x +x^2}}+\tan^{-1}\left(\frac{\sqrt{2}x}{1-x^2}\right) \cr \frac{df}{dx}&=\mathord? \end{align} $$
The answer is $$\frac{2\sqrt{2}}{1+x^4}$$
Hints: $$\frac{d\left(\log(1+x^2\pm\sqrt{2}x\right)}{dx}=\frac{2x\pm\sqrt{2}}{1+x^2\pm\sqrt{2}x}$$ and $$\left(1+x^2+\sqrt{2}x\right)\left(1+x^2-\sqrt{2}x\right)=\left(1+x^2\right)^2-\left(\sqrt{2}x\right)^2$$
Similarly,
$$\frac{d\left(\tan^{-1}\left(\frac{\sqrt{2}x}{1-x^2}\right)\right)}{dx}=\frac{1}{1+\left(\frac{\sqrt{2}x}{1-x^2}\right)^2}\frac{d\left(\frac{\sqrt{2}x}{1-x^2}\right)}{dx}$$
$$=\frac{1}{1+\left(\frac{\sqrt{2}x}{1-x^2}\right)^2}\frac{\sqrt{2}}{2}\left(\frac{1}{\left(1-x\right)^2}+\frac{1}{\left(1+x\right)^2}\right)$$
The rest is by calculations.