Let $f(t)=\frac{t}{1-t}$, $t \not= 1$.  If $y=f(x)$, then $x$ can be expressed as 
$\textbf{(A)}\ f\left(\frac{1}{y}\right)\qquad \textbf{(B)}\ -f(y)\qquad \textbf{(C)}\ -f(-y)\qquad \textbf{(D)}\ f(-y)\qquad \textbf{(E)}\ f(y)$
Since we know that $y=f(x)$, we can solve for $y$ in terms of $x$. This gives us
$y=\frac{x}{1-x}$
$\Rightarrow y(1-x)=x$
$\Rightarrow y-yx=x$
$\Rightarrow y=yx+x$
$\Rightarrow y=x(y+1)$
$\Rightarrow x=\frac{y}{y+1}$
Therefore, we want to find the function with $y$ that outputs $\frac{y}{y+1}$ Listing out the possible outputs from each of the given functions we get
$f\left(\frac{1}{y}\right)=\frac{1}{y-1}$
$f(y)=\frac{1}{1-y}$
$f(-y)=\frac{-y}{y+1}$
$-f(y)=\frac{y}{y-1}$
$-f(-y)=\frac{y}{y+1}$
Since $-f(-y)=\frac{y}{y+1}=x$ the answer must be $\boxed{C}$.