The problem
Let $p$ be defined by
$$ \begin{array}{c r c l} p : & \mathbb{R}^2 & \rightarrow & \mathbb{R}^3 \\ & (\theta,\phi) & \mapsto & \left( \frac{\cos \phi}{\cosh \theta},\frac{\sin \phi}{\cosh \theta},\tanh \theta \right) \end{array} $$
Show that $p$ is an immersion on the sphere s.t $(\theta,\phi)$ is not onto the poles.
My works
I know I have to prove that the differential of $p$ is injective. Then, I show that the jacobian matrix of $p$ is
$$ \mathrm{Jac}_{(\theta,\phi)}(p) = \left( \begin{array}{c c} - \frac{\cos \phi \sinh \theta}{\cosh^2 \theta} & - \frac{\sin \phi }{\cosh \theta} \\ - \frac{\cos \phi \sinh \theta}{\cosh^2 \theta} & \frac{\cos \phi }{\cosh \theta} \\ \frac{1}{\cosh^2 \theta} & 0 \\ \end{array} \right) $$
Next, I have to find that the rank of this jacobian matrix is equal to $0$. But I don't know how to do it. Do we have to determine the determinant? or should I show that the jacobian matrix is equal to $(0,0,0)$ if and only if $(\theta,\phi) = (0,0)$?
Thank you
$\cosh$ is everywhere nonzero, so calling the two columns $v_1$ and $v_2$, if we have $$ av_1 + bv_2 = 0, $$ then $\frac{a}{\cosh^2 \theta} + b \cdot 0 = 0$, so $a = 0$. Now look at $v_2$: its first two entries are just $(-\sin \phi, \cos \phi)$, multiplied by a nonzero number. These can never be simultaneously zero because $\sin^2 \phi + \cos^2 \phi = 1$ for all $\phi$.
In short, if a combination of the columns is zero, then the coefficients are both zero; hence the columns are linearly independent, hence the rank is $2$. That's what you need to show the function's an immersion. (Note that I did not show that the rank of the Jacobian is zero, as you asked, because (a) it's false, and (b) it's not what's needed anyhow.)