Answer
$f$ has no zeros.
Work Step by Step
To find the zeros of $f(x)=\frac{1}{(x-1)^2+1}$, we set
$$\frac{1}{(x-1)^2+1}=0 $$ which has no solutions -- that is, the function $f$ has no zeros.
Since $f(-x)\neq -f(x)$ nor $f(-x)\neq f(x)$ then the function is not symmetric about the origin nor about the y-axis and it is increasing on the interval $(-\infty,1)$. See the figure below.
