$$y=\sqrt[4]{\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}}\tag{1}$$
$$=\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{\frac{1}{4}}$$
$$\frac{dy}{dx}=\frac{d}{dx}\left(\sqrt[4]{\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}}\right)\tag{2}$$
$$u:=\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}~~\leftarrow~~\text{Preparation for chain rule}\tag{3}$$
$$\frac{dy}{dx}=\frac{d}{dx}\left(\sqrt[4]{u}\right)$$
$$=\frac{d}{du}\left(\sqrt[4]{u}\right)\frac{du}{dx}$$
$$=\frac{d}{du}\left(u^{\frac{1}{4}}\right)\frac{du}{dx}$$
$$=\frac{1}{4}u^{-\frac{3}{4}}\frac{du}{dx}$$
$$=\frac{1}{4}\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{-\frac{3}{4}}\frac{d}{dx}\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)$$
$$f\left(x\right):=\frac{d}{dx}\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)$$
$$g\left(x\right):=\left(a+x\right)\left(b+x\right)$$
$$h\left(x\right):=\left(a-x\right)\left(b-x\right)$$
$$\therefore~~f\left(x\right)=\frac{g'\left(x\right)h\left(x\right)-g\left(x\right)h'\left(x\right)}{h^{2}\left(x\right)}$$
$$g'\left(x\right)=\left(b+x\right)+\left(a+x\right)=a+b+2x$$
$$h'\left(x\right)=\left(-1\right)\left(b-x\right)+\left(-1\right)\left(a-x\right)$$
$$=\left(-1\right)\left(\left(b-x\right)+\left(a-x\right)\right)$$
$$=-\left(a+b-2x\right)$$
$$f\left(x\right)=\frac{\left(a+b+2x\right)\left(a-x\right)\left(b-x\right)-\left(a+x\right)\left(b+x\right)\left(-1\right)\left(a+b-2x\right)}{\left(a-x\right)^{2}\left(b-x\right)^{2}}$$
$$=\frac{\left(a+b+2x\right)\left(a-x\right)\left(b-x\right)+\left(a+x\right)\left(b+x\right)\left(a+b-2x\right)}{\left(a-x\right)^{2}\left(b-x\right)^{2}}=\frac{du}{dx}$$
$$\frac{dy}{dx}=\frac{1}{4}\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{-\frac{3}{4}}\underbrace{\frac{d}{dx}\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)}_{=\frac{du}{dx}}$$
$$=\frac{1}{4}\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{-\frac{3}{4}}\underbrace{\frac{\left(a+b+2x\right)\left(a-x\right)\left(b-x\right)+\left(a+x\right)\left(b+x\right)\left(a+b-2x\right)}{\left(a-x\right)^{2}\left(b-x\right)^{2}}}_{=:\text{green}\left(x\right)}$$
$$=\underbrace{\frac{1}{2}\sqrt[4]{\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}}\left(\frac{a}{a^{2}-x^{2}}+\frac{b}{b^{2}-x^{2}}\right)}_\text{What I want to finally derive. }$$
$$=\frac{1}{2}\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{\frac{1}{4}}\left(\frac{a}{a^{2}-x^{2}}+\frac{b}{b^{2}-x^{2}}\right)$$
$$=\frac{1}{2}\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{\frac{1}{4}}\left(\frac{a}{a^{2}-x^{2}}+\frac{b}{b^{2}-x^{2}}\right)\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{\frac{4}{4}}\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{-\frac{4}{4}}$$
$$=\frac{1}{2}\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{-\frac{3}{4}}\left(\frac{a}{a^{2}-x^{2}}+\frac{b}{b^{2}-x^{2}}\right)\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{\frac{4}{4}}$$
$$=\frac{1}{2}\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{-\frac{3}{4}}\left(\frac{a}{a^{2}-x^{2}}+\frac{b}{b^{2}-x^{2}}\right)\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{}\frac{2}{2}$$
$$=\frac{1}{4}\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{-\frac{3}{4}}\underbrace{\frac{2}{1}\left(\frac{a}{a^{2}-x^{2}}+\frac{b}{b^{2}-x^{2}}\right)\left(\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\right)^{}}_{\text{I want to derive this from}~\text{green}\left(x\right)}$$
$$\text{green}\left(x\right)=\frac{\left(a+b+2x\right)\left(a-x\right)\left(b-x\right)+\left(a+x\right)\left(b+x\right)\left(a+b-2x\right)}{\left(a-x\right)^{2}\left(b-x\right)^{2}}$$
$$=\frac{1}{\left(a-x\right)^{2}\left(b-x\right)^{2}}\left\{\left(a+b+2x\right)\left(a-x\right)\left(b-x\right)+\left(a+x\right)\left(b+x\right)\left(a+b-2x\right)\right\}$$
$$=\frac{1}{\left(a-x\right)\left(b-x\right)}\frac{1}{\left(a-x\right)\left(b-x\right)}\left\{\left(a+b+2x\right)\left(a-x\right)\left(b-x\right)+\left(a+x\right)\left(b+x\right)\left(a+b-2x\right)\right\}$$
$$=\frac{1}{\left(a-x\right)\left(b-x\right)}\left\{\left(a+b+2x\right)+\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\left(a+b-2x\right)\right\}$$
$$=\frac{1}{\left(a-x\right)\left(b-x\right)}\left\{\left(a+b+2x\right)+\frac{\left(a+x\right)\left(b+x\right)}{\left(a-x\right)\left(b-x\right)}\frac{\left(a+x\right)}{\left(a+x\right)}\left(a+b-2x\right)\right\}$$
$$=\frac{1}{\left(a-x\right)\left(b-x\right)}\left\{\left(a+b+2x\right)+\frac{\left(a+x\right)^{2}\left(b+x\right)}{\left(a^{2}-x^{2}\right)\left(b-x\right)}\left(a+b-2x\right)\right\}$$
$$=\frac{1}{\left(a-x\right)\left(b-x\right)}\left\{\left(a+b+2x\right)+\frac{\left(a+x\right)^{2}\left(b+x\right)}{\left(a^{2}-x^{2}\right)\left(b-x\right)} \frac{\left(b-x\right)^{-1}}{\left(b-x\right)^{-1}} \left(a+b-2x\right)\right\}$$
$$=\frac{1}{\left(a-x\right)\left(b-x\right)}\left\{\left(a+b+2x\right)+\frac{\left(a+x\right)^{2}\left(b+x\right)\left(b-x\right)^{-1}}{\left(a^{2}-x^{2}\right)} \left(a+b-2x\right)\right\}$$
$$=\frac{1}{\left(a-x\right)\left(b-x\right)}\left\{\left(a+b+2x\right)+\frac{\left(a+x\right)^{}\left(b+x\right)^{}\left(a+x\right)\left(b-x\right)^{-1}}{\left(a^{2}-x^{2}\right)} \left(a+b-2x\right)\right\}$$
Stucked.
BTW the derivation of name of green function is that I had painted the eqn and text with green in markdown however somehow MSE editor can't compile it correctly so I just removed that paining code .
