If $x \in(-\sqrt[3]{2}, 0]$, it is evidently not a solution, so let us assume otherwise. Then, we find $$f(x)=\frac{\sqrt{x^{4}+2 x}-x^{2}}{2 x}$$ which implies that $x f(x)^{2}+x^{2} f(x)-1 / 2=0$, by reverse engineering the quadratic formula. Therefore, if $x>0, f(x)$ is the unique positive real $t$ so that $x t^{2}+x^{2} t=1 / 2$. However, then $x$ is the unique positive real so that $x t^{2}+x^{2} t=1 / 2$, so $f(t)=x$. This implies that if $x>0$, then $f(f(x))=x$. Suppose that $f^{10}(x)=1$. Then, since $f(x)>0$, we find that $f(x)=f^{10}(f(x))=f^{11}(x)=f(1)$. Conversely, if $f(x)=f(1)$, then $f^{10}(x)=f^{9}(f(x))=f^{9}(f(1))=1$, so we only need to solve $f(x)=f(1)$. This is equivalent to $x^{2}+\sqrt{x^{4}+2 x}=1+\sqrt{3} \Longleftrightarrow \sqrt{x^{4}+2 x}=1+\sqrt{3}-x^{2} \Longrightarrow x^{4}+2 x=x^{4}-2(1+\sqrt{3}) x^{2}+(1+\sqrt{3})^{2}$, which is equivalent to $$2(1+\sqrt{3}) x^{2}+2 x-(1+\sqrt{3})^{2}=0$$ Obviously, if $x=1$ then $f(x)=f(1)$, so we already know 1 is a root. This allows us to easily factor the quadratic and find that the other root is $-\frac{1+\sqrt{3}}{2}$. This ends up not being extraneous-perhaps the shortest way to see this is to observe that if $x=-\frac{1+\sqrt{3}}{2}$, $$1+\sqrt{3}-x^{2}=(1+\sqrt{3})\left(1-\frac{1+\sqrt{3}}{4}\right)>0$$ so since we already know $$x^{4}+2 x=\left(1+\sqrt{3}-x^{2}\right)^{2}$$ we have $$\sqrt{x^{4}+2 x}=1+\sqrt{3}-x^{2}$$ Therefore, the sum of solutions is $\frac{1-\sqrt{3}}{2}$.