$(a) \Rightarrow (b)$:
$f(xy) = (xy)^{-1} = y^{-1}x^{-1} = x^{-1}y^{-1} = f(x)f(y)$
So $f$ is indeed a homomorphism.
$(b) \Rightarrow (a)$, or really not$(a) \Rightarrow$ not$(b)$:
Let $x$ and $y$ be two elements which don't commute. Then
$f(xy) = (xy)^{-1} = y^{-1}x^{-1}$ but $f(x)f(y) = x^{-1}y^{-1}$
Since $x$ and $y$ don't commute, their inverses can't commute either, since we would then have
$xy = ((xy)^{-1})^{-1} = (y^{-1}x^{-1})^{-1} = (x^{-1}y^{-1})^{-1} = ((yx)^{-1})^{-1} = yx$
$(a) \Rightarrow (c)$:
$f(xy) = (xy)(xy) = xyxy = x^2y^2 = f(x)f(y)$
Not$(a) \Rightarrow$ not$(c)$:
Let $x$ and $y$ be two elements that don't commute. We then have
$f(xy) = (xy)(xy) = xyxy$ but $f(x)f(y) = x^2y^2 = xxyy$
These cannot be equal to each other, since we would then have the contradiction
$xyxy = xxyy \Rightarrow x^{-1}xyxyy^{-1} = x^{-1}xxyyy^{-1} \Rightarrow yx = xy$
Hence both $(b)$ and $(c)$ are equivalent to $(a)$.