Answer
$Area\left( S \right) = \pi $

Work Step by Step
From ${z^2} = {x^2} + {y^2}$ and ${y^2} + {z^2} \le 1$ we get
${y^2} + {z^2} = {x^2} + 2{y^2} \le 1$
So, we can consider the domain ${\cal D}$ to be described by
${\cal D} = \left\{ {\left( {x,y} \right):{x^2} + 2{y^2} \le 1} \right\}$
We notice that the boundary of ${\cal D}$ is an ellipse given by
${x^2} + {\left( {\dfrac{y}{{1/\sqrt 2 }}} \right)^2} = 1$,
where the semimajor axis and the semiminor axis are $a=1$ and $b = \dfrac{1}{{\sqrt 2 }}$, respectively.
We can parametrize the surface by $G\left( {x,y} \right) = \left( {x,y, \pm \sqrt {{x^2} + {y^2}} } \right)$. Since $z \ge 0$, we choose the parametrization $G\left( {x,y} \right) = \left( {x,y,\sqrt {{x^2} + {y^2}} } \right)$.
${{\bf{T}}_x} = \dfrac{{\partial G}}{{\partial x}} = \left( {1,0,\dfrac{x}{{\sqrt {{x^2} + {y^2}} }}} \right)$
${{\bf{T}}_y} = \dfrac{{\partial G}}{{\partial y}} = \left( {0,1,\dfrac{y}{{\sqrt {{x^2} + {y^2}} }}} \right)$
${\bf{N}}\left( {x,y} \right) = {{\bf{T}}_x} \times {{\bf{T}}_y} = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
1&0&{\dfrac{x}{{\sqrt {{x^2} + {y^2}} }}}\\
0&1&{\dfrac{y}{{\sqrt {{x^2} + {y^2}} }}}
\end{array}} \right|$
${\bf{N}}\left( {x,y} \right) = - \dfrac{x}{{\sqrt {{x^2} + {y^2}} }}{\bf{i}} - \dfrac{y}{{\sqrt {{x^2} + {y^2}} }}{\bf{j}} + {\bf{k}}$
$||{\bf{N}}\left( {x,y} \right)|| = \sqrt {{{\left( { - \dfrac{x}{{\sqrt {{x^2} + {y^2}} }}} \right)}^2} + {{\left( { - \dfrac{y}{{\sqrt {{x^2} + {y^2}} }}} \right)}^2} + 1} $
$||{\bf{N}}\left( {x,y} \right)|| = \sqrt {\dfrac{{2{x^2} + 2{y^2}}}{{{x^2} + {y^2}}}} = \sqrt 2 $
By Theorem 1, the surface area of $S$ is
$Area\left( S \right) = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} ||{\bf{N}}\left( {x,y} \right)||{\rm{d}}x{\rm{d}}y$
$Area\left( S \right) = \sqrt 2 \mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}x{\rm{d}}y$
Notice that $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_{\cal D}^{} {\rm{d}}x{\rm{d}}y$ is the area of the ellipse ${\cal D} = \left\{ {\left( {x,y} \right):{x^2} + 2{y^2} \le 1} \right\}$, where the semimajor axis and the semiminor axis are $a=1$ and $b = \dfrac{1}{{\sqrt 2 }}$, respectively.
In Exercise 24 of Section 14.6 on page 753, we know that the area of the ellipse is $A = \pi ab$. So,
$Area\left( S \right) = \sqrt 2 \cdot \pi \cdot 1 \cdot \dfrac{1}{{\sqrt 2 }} = \pi $