If ${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\frac{3\pi }{2}$ and$f\left(1\right)=2,f(p+q)=f\left(p\right)·f\left(q\right),\forall p,q\in R$, then $x^{f\left(1\right)}+y^{f\left(2\right)}+z^{f\left(3\right)}-\frac{(x+y+z)}{x^{f\left(1\right)}+y^{f\left(2\right)}+z^{f\left(3\right)}}$ is equal to

$\because \frac{-\pi }{2}\le {\sin }^{-1}x\le \frac{\pi }{2},\frac{-\pi }{2}\le {\sin }^{-1}y\le \frac{\pi }{2}$

and $\frac{-\pi }{2}\le {\sin }^{-1}z\le \frac{\pi }{2}$

Given that, 

${\sin }^{-1}x+{\sin }^{-1}y+{\sin }^{-1}z=\frac{3\pi }{2}$

Which is possible only when

${\sin }^{-1}x={\sin }^{-1}y={\sin }^{-1}z=\frac{\pi }{2}$

$\Rightarrow x=y=z=1$

Put $p=q=1$

Then, $f\left(2\right)=f\left(1\right)f\left(1\right)=2\times 2=4$

and put $p=1,q=2$

then, $f\left(3\right)=f\left(1\right) f\left(2\right)$

$=2·2^2=8$

$\therefore x^{f\left(1\right)}+y^{f\left(2\right)}+z^{f\left(3\right)}-\frac{x+y+z}{x^{f\left(1\right)}+y^{f\left(2\right)}+z^{f\left(3\right)}}$

$=1+1+1-\frac{3}{1+1+1}=3-1=2$