There is an easier way.

We have,

$\displaystyle{ f(x) = a \log{\frac{1+x}{1-x}} +bx^3 + c \sin x +5}$

Taking $5$ to the Left hand side,

$\displaystyle{ f(x) - 5= a \log{\frac{1+x}{1-x}} +bx^3 + c \sin x }$

Let this be Statement 1.

Now putting $x=-x$ in the original equation,

$\displaystyle{ f(-x) = a \log{\frac{1-x}{1+x}} -bx^3 - c \sin x +5}$

Also,

$\displaystyle{ f(-x) = -a \log{\frac{1+x}{1-x}} -bx^3 - c \sin x +5}$

Using Statement 1,

$\displaystyle{ f(-x) = 5 - f(x)+5}$

Further simplifying,

$\displaystyle{ f(-x) = 10 - f(x)}$

Now putting $x=\log _{ 3 }{ 2 }$ we get $\boxed{f(\log _{ 3 }{ \frac{1}{2} })=6}$

Alternative Solution

Let $g(x)=f(x)-5$ .

Observe that $g(x)$ is an odd function.

Therefore $g(-x)=-g(x)$ .

$f(-x)-5=5-f(x)$ $f(-x)=10-f(x)$