gaaanz einfach :
$$ f(x) = \frac1{x+1} $$
$$ g(x)= x ^2 - 2x $$
$$ g(f(x))= f^2(x) - 2f(x) $$
$$ g(f(x))= \left(\frac1{x+1}\right)^2 - 2\frac1{x+1} $$
$$ g(f(x))= \frac1{\left(x+1\right)^2} - \frac2{x+1} $$
$$ g= \frac1{\left(x+1\right)^2} -2 \frac{x+1} {\left(x+1\right)^2} $$
$$ g= \frac{1-2x-2} {\left(x+1\right)^2} $$
$$ g= \frac{-2x-1} {\left(x+1\right)^2} $$
$$ g {\left(x+1\right)^2}= - 2x-1$$
$$ g {\left(x^2+2x+1\right)}= - 2x-1$$
$$ g {\left(x^2+2x+1\right)}+ 2x+1=0$$
$$ g \cdot x^2+2(g+1) \cdot x+2=0$$
$$x_{1,2}= \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
$$x_{1,2}= \frac{-2(g+1)\pm \sqrt{4(g+1)^2-4g\cdot 2}}{2g}$$
$$x_{1,2}= \frac{-2(g+1)\pm \sqrt{4(g^2+2g+1)-8g}}{2g}$$
$$x_{1,2}= \frac{-2(g+1)\pm \sqrt{4g^2+8g+4-8g}}{2g}$$
$$x_{1,2}= \frac{-2(g+1)\pm \sqrt{4g^2+4}}{2g}$$
$$x_{1,2}= \frac{-2(g+1)\pm 2\sqrt{g^2+1}}{2g}$$
$$x_{1,2}= \frac{-(g+1)\pm \sqrt{g^2+1}}{g}$$
$$x_{1,2}= -\frac gg-\frac 1g\pm \sqrt{\frac{g^2+1}{g^2}}$$
$$x_{1,2}= -1-\frac 1g\pm \sqrt{1+\frac{1}{g^2}}$$
