smc

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 $\boxed{-f(-y)}$.