Define $$\tag{1} \begin{equation} S:=\{(x, y, z) \in A \times \mathbb{R} \mid z=h(x, y)\}, \end{equation} $$ where $A = \begin{equation} \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \times\left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \end{equation}$ and $h(x,y):=\log(\frac{\cos y}{\cos x})$. Further, define $F : A\rightarrow\mathbb{R}^3$ by $$\tag{2} F(x,y)=(x,y,h(x,y)). $$ Now, I am supposed to show that $F(A)=S$ and that the smooth map $F$ is bijective with continuous inverse $$\tag{3} \begin{equation} F^{-1}(x, y, z)=(x, y) \end{equation}. $$ To show that $F(A)=S$, I need to show that $\forall s\in S\exists (x,y)\in A$ such that $F(x,y)=S$. To show that it is a bijection, I need to show that $(x,y)$ in unique. So pick some $(x,y,z)\in S$. Then $$\tag{4} F(x,y)=(x,y,h(x,y))=(x,y,z) $$ Hence, $F(A)=S$. It also shows that $(x,y)$ is unique, because if $(x',y')\ne (x,y)$, then $(x,y,z)\ne(x',y',z')$. Finally, it also shows that eq. $(3)$ holds. Since any smooth function is continuous, and the inverse of a continuous is continuous, we know that $F^{-1}$ in eq. $(3)$ is continuous.
Did I misunderstand something, or is eq. $(4)$ really sufficient to solve the given problem?