You can compute the derivative directly by viewing the spheres as submanifolds in $\mathbb H$ and the space of purely imaginary quaternions, respectively. Indeed, the map $f(z)=ziz^*$ makes sense as a map from $\mathbb H$ to imaginary quaternions, so you can differentiate it as any map $\mathbb R^4\to\mathbb R^3$, and then think about the restriction to the unit spheres. In partiuclar, you can compute $Df(z)(w)$ as the derivative at $t=0$ of $f(z+tw)$. Oberserving that quaternionic multiplication is $\mathbb R$ bilinear, while conjugation is $\mathbb R$-linear, you get $Df(z)(w)=wiz^*+ziw^*$, and this is twice the imaginary part of $wiz^*$.
So what you actually have to prove is that if you have a purely imaginary quaternion $q$ perpendicular to $ziz^*$ (i.e. tangent to $S^2$ in that point), then you can write it as $wiz^*+ziw^*$, for some $w$ which is perpendicular to $z$ (and hence tangent to $S^3$ in $z$). Now this is easy for $z=1$. Here $q$ must be a linear combination of $j$ and $k$, while $w$ has to be purely imaginary, and you immediately write out an explicit solution. For general $z$, the orthogonal space to $ziz^*$ is spanned by $zjz^*$ and $zkz^*$ and you can similarly find an explicit solution.
Having verified that $\pi$ is a submersion, it follows that it is surjective. Since $d\pi(z)$ is surjective, the implicit function theorem shows that the image of $\pi$ contains an open nighborhood of $z$ in $S^2$ and since this works for each $z$, you see that $\pi(S^3)$ is open in $S^2$. On the other hand, continuity of $\pi$ shows that $\pi(S^3)$ is a compact subset of $S^2$ and thus closed. Since $S^2$ is connected, you get $\pi(S^3)=S^2$.
