How to solve for x from$$ \frac{1}{\sqrt{1-x^2/m^2}}=\frac{1}{\sqrt{1-y^2/m^2}} \cdot\frac{1}{\sqrt{1-z^2/m^2}}\cdot(1+\dfrac{yz}{m^2}) $$ to $$ x = \frac{y+z}{1+yz/m^2}$$
There is also a tip, which says $$\dfrac{x^2}{m^2}=\frac{\left(\frac{1}{\sqrt{1-x^2/m^2}}\right)^2-1}{\left(\frac{1}{\sqrt{1-x^2/m^2}}\right)^2}$$
I took both sides to the $-2$ power: $$1-\frac{x^2}{c^2}=\frac{\left(1-\frac{y^2}{c^2}\right)\left(1-\frac{z^2}{c^2}\right)}{\left(1+\frac{yz}{c^2}\right)^2}$$ Then took $1$ minus both sides: $$\begin{align}\frac{x^2}{c^2}&=\frac{\left(1+\frac{yz}{c^2}\right)^2-\left(1-\frac{y^2}{c^2}\right)\left(1-\frac{z^2}{c^2}\right)}{\left(1+\frac{yz}{c^2}\right)^2}\\ &=\frac{\color{red}{1}+\frac{2yz}{c^2}\color{blue}{+\frac{y^2z^2}{c^4}}\color{red}{-1}+\frac{y^2}{c^2}+\frac{z^2}{c^2}\color{blue}{-\frac{y^2z^2}{c^4}}}{\left(1+\frac{yz}{c^2}\right)^2}\\ &=\frac{\frac{(y+z)^2}{c^2}}{\left(1+\frac{yz}{c^2}\right)^2}\end{align}$$ The terms with corresponding $\color{red}{\text{c}}\color{orange}{\text{o}}\color{yellow}{\text{l}}\color{green}{\text{o}}\color{blue}{\text{r}}\color{violet}{\text{s}}$ canceling. Then multiply by $c^2$ and take square root to get $$x=\frac{y+z}{1+\frac{yz}{c^2}}$$ Is that what you wanted, or did you rather want to start with the last expression for $x$ and substitute into $\gamma=\left(1-\frac{x^2}{c^2}\right)^{-1/2}$?