Answer
$f\left(\frac{1}{1-a}\right)=\frac{a}{a-a^2+1}$
Work Step by Step
Evaluate the function at $x=1/(1-a)$, in $f(x)=\frac{a x}{a+x}$.
$$
f\left(\frac{1}{1-a}\right)=\frac{a \frac{1}{1-a}}{a+\frac{1}{1-a}}=\frac{\frac{a}{1-a}}{\frac{a(1-a)+1}{1-a}}=\frac{a}{1-a} \cdot \frac{1-a}{a-a^2+1}=\frac{a}{a-a^2+1}
$$
