Page 70, Bott and Tu, Differential Form in Algebraic Topology

Question

I am trying to solve the second part of exercise 6.32. I do not understand how the form $d\phi$ is smooth on $S^2$. I am trying to express $d\phi$ in terms of dx,dy,dz. Even if I use the standard coordinate chart on $S^2$ given by $\psi: B(0,1) \rightarrow S^2; (x,y)\mapsto (x,y,\sqrt{1-x^2-y^2})$ ,then the expression for $\phi$ becomes $\arccos(\sqrt{1-x^2-y^2})$ . Thus $d\phi=\frac{x}{\sqrt{1-x^2-y^2} \sqrt{x^2+y^2}}dx + \frac{y} {\sqrt{1-x^2-y^2}\sqrt{x^2+y^2}}dy$ . I can not see why it is smooth at the point (0,0) or $\psi(0,0)=(1,0,0)$. Kindly help me.