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 $S_r(0)$?

Question

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 $r$?

Consider the function $H:\mathbb R^3 \to \mathbb R$, $H(x,y,z)=x^2+y^2+z^2$

Claim:F is transverse to $S_r(0)$ if and only if $r^2$ is a regular value of $H\circ G$. Why is this true?

I need that $\forall x \in F^{-1}(S_r(0)), dF_{x}(T_x\mathbb R^2)$ together with $T_{F(x)}S_r(0)$ to span the whole $R^3$, Note that $T_{F(x)}S_r(0)$ is of dimension $2$, then I only need $dF_{x}(T_x\mathbb R^2)$ to have a nonzero vector which does not belong to $T_{F(x)}S_r(0)$, but how should I go on to prove the claim?