1. We are given $f(x)=\ln\left(\frac{1-x}{1+x}\right), \quad |x|<1.$

We need to find $f\left(\frac{2x}{1+x^2}\right).$

2. Substitute $t=\dfrac{2x}{1+x^2}$ into the definition of $f$: $f\left(\frac{2x}{1+x^2}\right)=\ln\left(\frac{1-\frac{2x}{1+x^2}}{1+\frac{2x}{1+x^2}}\right).$

3. Simplify numerator and denominator inside the logarithm: $1-\frac{2x}{1+x^2}=\frac{1+x^2-2x}{1+x^2}=\frac{(1-x)^2}{1+x^2},$ $1+\frac{2x}{1+x^2}=\frac{1+x^2+2x}{1+x^2}=\frac{(1+x)^2}{1+x^2}.$

So, $\frac{1-\frac{2x}{1+x^2}}{1+\frac{2x}{1+x^2}}=\frac{\frac{(1-x)^2}{1+x^2}}{\frac{(1+x)^2}{1+x^2}}=\frac{(1-x)^2}{(1+x)^2}.$

4. Therefore, $f\left(\frac{2x}{1+x^2}\right)=\ln\left(\frac{(1-x)^2}{(1+x)^2}\right).$

Using $\ln(a^2)=2\ln|a|$, and since $|x|<1$, we have $1-x>0$ and $1+x>0$, so $\ln\left(\frac{(1-x)^2}{(1+x)^2}\right)=2\ln\left(\frac{1-x}{1+x}\right).$

Hence, $f\left(\frac{2x}{1+x^2}\right)=2f(x).$

5. Now evaluate the options:

• A: $2f(x^2)$ — not equal in general.
• B: $2f(x)$ — correct.
• C: $(f(x))^2$ — incorrect.
• D: $-2f(x)$ — incorrect.

Therefore, the correct answer is B.