Let $F:\mathbb{R}^2 \to \mathbb{R}^3$ be given by $F(x,y)=(e^y\cos x,e^y\sin x,e^{-y})$. For which $r>0$ is $F$ transverse to the sphere $S_r(0)$ which center at the origin and the radius is $r$?
My try:
$$dF(x,y)=\begin{bmatrix} -e^y\sin x &e^y\cos x \\ e^y\cos x & e^y\sin x \\ 0 & -e^{-y} \\ \end{bmatrix}$$ Since $S_r \subset \mathbb{R^3}$ is an embedded manifold, $F$ is transverse to $s_r$ if for every $(x,y) \in F^{-1}(S_r)$, $T_{F(x,y)}S_r$ and $dF_{(x,y)}(T_{(x,y)}\mathbb{R^2})$ together span $T_{F(x,y)}\mathbb{R^3}$ which is equivalent to show that for every $(x,y) \in F^{-1}(S_r), T_{F(x,y)}S_r$ and $dF_{(x,y)}(\mathbb{R^2})$ span $\mathbb{R^3}$.
Now $$(x,y) \in F^{-1}(S_r)\iff e^{2y}+e^{-2y}=r^2 \iff y=\frac{1}{2}\ln\left(\frac{r^2 \pm\sqrt{r^4-4} }{2}\right)$$
For $r \lt \sqrt{2}$, $y$ doesn't exist. Let $(x,y) \in F^{-1}(S_r)$. Then suppose that $(a,b,c) \in T_{F(x,y)}S_r$ i.e $ae^{y}\cos x + be^y \sin x + ce^{-y}=0$.
$T_{F(x,y)}S_r$ and $dF_{(x,y)}(\mathbb{R^2})$ will span $\mathbb{R^3} \iff (a,b,c),(-e^y \sin x,e^y\cos x,0)$ and $(e^y\cos x,e^y\sin x,-e^{-y})$ are linearly independent iff the determinant of these three vectors is non-zero.
To this end, Let's evaluate the following:
$$\det\begin{vmatrix} -e^y \sin x &e^y\cos x & 0\\e^y\cos x & e^y\sin x & -e^{-y}\\a & b & c\end{vmatrix}=0 \iff c(e^{2y}-e^{-2y})=0$$
Well if $y=0$ then, we are lead to the case of $r=\sqrt{2}$ and this kind of makes me believe that $F$ is transverse for $r \gt \sqrt{2}$. But I am unable to deal with the case when $c=0$. How do I imply the conclusion from here?
Thanks for the help!!
Here's a much easier approach. The two surfaces fail to be transverse at $(a,b,c)$ if and only if their normal vectors at $(a,b,c)$ are parallel. So, let's consider the point $F(x,y)$. The normal vector to the image of $F$ at that point is $-(\cos x,\sin x, e^{2y})$. But the normal vector to the sphere through that point is a scalar multiple of $F(x,y)$ itself. So we are asking when $$(\cos x,\sin x,e^{2y}) = \lambda (e^y\cos x,e^y\sin x,e^{-y})$$ for some scalar $\lambda$. It is easy to see that this happens if and only if $e^{2y} = e^{-2y}$, so $y=0$. But $F(x,0)$ lies on the sphere of radius $\sqrt2$ for all $x$.
This seems to agree with your partial result, but I don't have to worry about anything else. :)