The system in the symmetric form is given by $$\frac{dx}{x^2-y^2-z^2}=\frac{dy}{2xy}=\frac{dz}{2xz}.$$ Rewrite using the derivatives $$\frac{dx}{dt}=x^2-y^2-z^2,$$ $$\frac{dy}{dt}=2xy,$$ $$\frac{dz}{dt}=2xz.$$
I think that $$\frac{dy}{dt}=\frac{dz}{dt}$$ $$xy^2=xz^2+C1$$ $$\frac{dx}{x^2-y^2-z^2}=\frac{d(y+z)}{2x(y+z)}$$ I do not know what's next may all my steps do not correct?
From $\frac{dy}{2xy}=\frac{dz}{2xz}$, you have $$ \frac{dy}{y}=\frac{dz}{z} $$ and hence $y=c_1z$. Put this in $\frac{dx}{x^2-y^2-z^2}=\frac{dz}{2xz}$ and then you have $$\frac{dx}{x^2-(c_1^2+1)z^2}=\frac{dz}{2xz}$$ which can be written as $$ \frac{dx}{dz}=\frac{x^2-(c_1^2+1)z^2}{2xz}.$$ The above equation is a homogeneous DE and you can solve by setting $x=uz$. I omit the detail.