Domain:
f(x) has a denominator, which can't be zero. So, its domain is given by
![x^2-1\\eq 0 \iff x^2\\eq 1 \iff x\\eq \pm 1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/wvtcg88r0gcokjoapls9iksuibxja4u2ym.png)
g(x) has a denominator as well. Moreover, it has a root. So, the content of the root can't be negative:
![x-1\geq 0 \iff x \geq 1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/qsvwq4emninylp2kwyxuqfl3fvc0nfp8x2.png)
And the denominator can't be zero:
![√(x-1)\\eq 0 \iff x-1 \\eq 0 \iff x \\eq 1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/sdmz0x53kfikeam6qguccyy23wab0kqo4b.png)
So, the domain is
![x>1](https://img.qammunity.org/2020/formulas/mathematics/middle-school/ko3a2iufko3s5odrr03oqu10d47yyva6i5.png)
Composition:
We have
![f(g(x))=(g^2(x))/(g^2(x)-1) = ((1)/(x-1))/((1)/(x-1)-1) = ((1)/(x-1))/((1-x+1)/(x-1)) = ((1)/(x-1))/((2-x)/(x-1))=(1)/(2-x)](https://img.qammunity.org/2020/formulas/mathematics/middle-school/axmtiavdr7h50dcag4ey3sfopu32itk0im.png)
The domain of this function is
![2-x\\eq 0 \iff x\\eq 2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/hprer3ara63mfjjwob11oic6cel5mpkg2o.png)
But we also have to remember about the domain of g(x): if g(x) is undefined, we can't compute f(g(x))!
So, the domain of f(g(x)) is
![x>1\ \land x\\eq 2](https://img.qammunity.org/2020/formulas/mathematics/middle-school/83epwu57b83htupnxp8n93uh3ej1zjelwk.png)