Have fun with functions!

$f(x)^2 \times f\left(\dfrac{1-x}{1+x}\right)=x^3$ $\quad ...(1)$

Substituting $\dfrac{1-x}{1+x}$ into $(1)$ we get :

$f\left(\dfrac{1-x}{1+x}\right)^2 \times f(x)=\dfrac{1-x}{1+x}$ $\quad ....(2)$

Dividing both the equations we get:

$\dfrac{f(x)}{f\left(\dfrac{1-x}{1+x}\right)}=\left(\dfrac{x(1+x)}{1-x} \right)^3$

or $f\left(\dfrac{1-x}{1+x}\right)=\left(\dfrac{1-x}{x(1+x)} \right)^3 \times f(x)$

Replacing this value into $(1)$ we get:

$f(x)^3 \times \left(\dfrac{1-x}{x(1+x)}\right)^3=x^3$

Taking cube root on both sides we get:

$f(x)=x^2 \left(\dfrac{1+x}{1-x}\right)$

Hence $f(10)=\dfrac{-1100}{9}$ and $a+b=\boxed{-1091}$ .