If $g(x)=1-x^2$ and $f(g(x))=\frac{1-x^2}{x^2}$ when $x\not=0$, then $f(1/2)$ equals 
$\mathrm{(A)\ } 3/4 \qquad \mathrm{(B) \ }1 \qquad \mathrm{(C) \  } 3 \qquad \mathrm{(D) \  } \sqrt{2}/2 \qquad \mathrm{(E) \  }\sqrt{2}$
Notice that if we can find a $y$ such that $g(y)=\frac{1}{2}$, then we can plug that into the second equation to get $f\left(\frac{1}{2}\right)=f(g(y))=\frac{1-y^2}{y^2}$. Thus, we let $1-y^2=\frac{1}{2}\implies y^2=\frac{1}{2}$. Notice that, since our final expression only involves $y^2$, we don't need to take the square root. Thus, $f\left(\frac{1}{2}\right)=\frac{1-y^2}{y^2}=\frac{1-\frac{1}{2}}{\frac{1}{2}}=1, \boxed{\text{B}}$.