smc

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 $1-y^2=\frac{1}{2}⟹y^2=\frac{1}{2}$. Notice that, since our final expression only involves $y^2$, we don't need to take the square root. Thus, $f\left(\frac{1}{2}\right)=\frac{1-y^2}{y^2}=\frac{1-\frac{1}{2}}{\frac{1}{2}}=1,\boxed{1}$.