Say $a_{2}=a$. Then using the recursion equation, we have $a_{3}=-1, a_{4}=\frac{a+1}{a-1}, a_{5}=\frac{-a+1}{a+1}, a_{6}=-\frac{1}{a}$, $a_{7}=-\frac{2 a}{a^{2}-1}$, and $a_{8}=1$ Now we have $a_{10} a_{8} a_{7}=a_{10}+a_{8}+a_{7}$. No value of $a_{10}$ can satisfy this equation iff $a_{8} a_{7}=1$ and $a_{8}+a_{7} \neq 0$. Since $a_{8}$ is 1, we want $1=a_{7}=-\frac{2 a}{a^{2}-1}$, which gives $a^{2}+2 a-1=0$. The only positive root of this equation is $\sqrt{2}-1$. This problem can also be solved by a tangent substitution. Write $a_{n}=\tan \alpha_{n}$. The given condition becomes $$\alpha_{n+2}+\alpha_{n}+\alpha_{n-1}=0$$ We are given $\alpha_{0}=0, \alpha_{1}=\pi / 4$, and $\alpha_{2} \in(0, \pi / 2)$. Using this, we can recursively compute $\alpha_{3}, \alpha_{4}, \ldots$ in terms of $\alpha_{2}$ until we get to $\alpha_{10}=\frac{3 \pi}{4}-2 \alpha_{2}$. For $a_{10}$ not to exist, we need $\alpha_{10} \equiv \pi / 2 \bmod \pi$. The only possible value of $\alpha_{2} \in(0, \pi / 2)$ is $\alpha_{2}=\pi / 8$, which gives $a_{2}=\tan \pi / 8=\sqrt{2}-1$.