Which parts of $S$ are not covered by these parametrizations? Prove that the set $S$ = {$(x,y,z) \in \Bbb{R}^3 | z = x^2 - y^2$ } is a differentiable surface and verify that 
$f(u,v) = (u + v, u - v, 4uv), (u, v) \in \Bbb{R}^3,$ and
$g(u,v) = (ucosh(v), usinh(v), u^2),  (u, v) \in \Bbb{R}^3, u \neq 0$,
are parametrizations of S. Which parts of $S$ are not covered by these parametrizations?
The question in black is my main concern, but I write my answer to show that I made some progress... 
$f, g $ are differentiable, because each of the component functions $f_i , g_i $
are continuous and therefore, differentiable.
Taking
$f(u,v) = (u + v, u - v, 4uv), (u, v) \in \Bbb{R}^3,$ I want to see that this is a parametrization, meaning, if I take a point in $f$ it should satisfy the equation defined in $S$:
Let $ x = u + v, y=u-v $. This implies that 
$u = \frac{x+y}{2}, v = \frac{x-y}{2}$. 
As $z = 4uv, $ then $z = 4uv = 4(\frac{x+y}{2})(\frac{x-y}{2})= x^2 - y^2$. This shows a pont given by the function $f$ is also in $S$.
Now, taking $g(u,v) = (ucosh(v), usinh(v), u^2),  (u, v) \in \Bbb{R}^3, u \neq 0$, let
$x= vcosh(u)$
$y = vsinh(u)$
$z= v^2$. Then $x^2 - y^2 = v^2 = z $, so $g$ is a parametrization of $S$.
I don't see which parts of S are not covered by these parametrizations...Maybe I would appreciate a more elaborate/math written hint than Which parts of $S$ these parametrization cover? Thanks
 A: First consider $f(u,v)$ . . .

Since the equation 
$$(u+v)^2 - (u-v)^2 = 4uv$$
is an identity, it follows that $f(u,v) \in S$, for all $u,v \in \mathbb{R}$.

Moreover, if $(x,y,z)$ is any point of $S$, then letting 
$u = {\large{\frac{x+y}{2}}}$
and 
$v = {\large{\frac{x-y}{2}}}$, we get
\begin{align*}
u + v &= {\small{\frac{x+y}{2}}} + {\small{\frac{x-y}{2}}}=x\\[4pt]
u - v &= {\small{\frac{x+y}{2}}} - {\small{\frac{x-y}{2}}}=y\\[4pt]
4uv &= 
4
\left(
{\small{\frac{x+y}{2}}}
\right)
\left(
{\small{\frac{x-y}{2}}
}\right)
=x^2-y^2=z\\[8pt] 
\implies\; f(u,v) &= (u+v,u-v,4uv) = (x,y,z)\\[4pt]
\end{align*}
Thus, all points of $S$ are realized by the parametrization $f(u,v)$. 

Hence, $f(u,v)$ covers $S$.

Next consider $g(u,v)$ . . .

Since the equation 
$$(v\cosh(u))^2- (v\sinh(u))^2 = v^2$$
is an identity, it follows that $g(u,v) \in S$, for all $u,v \in \mathbb{R}$.

But since the $z$-coordinate of $g(u,v) = u^2$ and $u \ne 0$, the parametrization only yields points in the set 
$$S^{+} = \{(x,y,z) \in S\mid z>0\}$$
In particular, the parametrizaton $g(u,v)$ doesn't cover all of $S$, but only some subset of $S^{+}$.

Claim it covers all of $S^{+}$.

Thus, suppose $(x,y,z) \in S$ with $z > 0$.
\begin{align*}
\text{Then}\;\;&x^2 - y^2 = z\\[4pt]
\implies\; &x^2 \ge z\\[4pt]
\implies\; &|x| \ge \sqrt{z}\;\;\text{and}\;\;|x| > 0\\[14pt]
\text{Let}\;\;&u = {\small{\frac{x}{|x|}}}\sqrt{z}\\[4pt]
&v = \sinh^{-1}\left({\small{\frac{y}{u}}}\right)\\[12pt]
\text{Then:}\;\;\;\;\,\\[4pt]
{\small{\bullet}}\;\;u\cosh(v) &= u\cosh\left(\sinh^{-1}\left({\small{\frac{y}{u}}}\right)\right)\\[4pt]
&=u\sqrt{\left({\small{\frac{y}{u}}}\right)^{\!2} + 1}\\[4pt]
&={\small{\frac{u}{|u|}}}\sqrt{y^2+u^2}\\[4pt]
&={\small{\frac{x}{|x|}}}\sqrt{y^2+z}&&
\text{[since $u ={\small{\frac{x}{|x|}}}\sqrt{z}\,$]}\\[4pt]
&={\small{\frac{x}{|x|}}}\sqrt{x^2}&&
\text{[since $z=x^2-y^2$]}\\[4pt]
&=x\\[12pt]
{\small{\bullet}}\;\;u\sinh(v) &= u\sinh\left(\sinh^{-1}\left({\small{\frac{y}{u}}}\right)\right)\\[4pt]
&= u\left({\small{\frac{y}{u}}}\right)\\[4pt]
&= y\\[12pt]
{\small{\bullet}}\;\;u^2 &= \left({\small{\frac{x}{|x|}}}\sqrt{z}\right)^{\!2}\\[4pt]
&=z\\[12pt]
\implies\; g(u,v) &= (u\cosh(v),u\sinh(v),u^2) = (x,y,z)\\[4pt]
\end{align*}
hence, as claimed, all points of $S^{+}$ are realized by the parametrization $g(u,v)$. 

Thus, $g(u,v)$ covers only $S^{+}$, not all of $S$.
