computation on hyper surface $z=x^2+y^2$ I have problem with following exercise

Consider the hypersurface $M$ parametrized by $z=x^2+y^2$. 
Endow this with the Riemannian metric induced from the $\mathbb{R}^3$. 
Compute the sectional curvature. 
 A: I'd like to know my calculation process is done correctly. 
Note $z=x^2+y^2$, in $\mathbb{R}^3$, by introducing spherical coordinate 
such that
\begin{align}
x=  r\cos{\theta}, \quad y= r\sin{\theta}  , \quad z=r^2
\end{align}
we have
\begin{align}
x^2 +y^2 =  r^2= z
\end{align}
and doing calculation with chain rule
\begin{align}
\partial_{r} = \cos(\theta)\partial_x + \sin(\theta) \partial_{y} +2r \partial_z\end{align}
\begin{align}
\partial_{\theta} = -r\sin(\theta) \partial_x + r\cos(\theta)  \partial_y 
\end{align}
Thus 
\begin{align}
ds^2 = (1+4r^2)dr^2 +  r^2d\theta^2
\end{align}
and sectional curvature 
How can i use sectional curvature formula for this case? 
I know 
Note i used the formula for sectional curvature for geodesic polar coordinate
$i.e$ 
\begin{align}
ds^2 =dr^2 +f(r,\theta)^2 d\theta^2 \qquad K=-\frac{1}{f}\frac{\partial^2 f}{\partial r^2}
\end{align}
A: This method was using  Monge surface parametrization, which was introduced by @THW. in my former question.  

Endowment of $\mathbb{R}^3$, we can do Gauss approach as follows.
\begin{align}
\mathbf{x}(r,\theta)=(f(r),r\cos(\theta), r\sin(\theta))
\end{align}
where $f(r)=r^2=z$.
\begin{align}
&\mathbf{x}(r,\theta)=\left(f(r),r\cos(\theta), r\sin(\theta)  \right) \quad \mathbf{x}_r = \left(f'(r), \cos(\theta), \sin(\theta) \right) \\
& \mathbf{x}_{rr} = \left( f''(r),0,0 \right) \quad \mathbf{x}_\theta = \left( 0,-r\sin(\theta), r\cos(\theta) \right) \\
&\mathbf{x}_{\theta\theta} = \left( 0,-r\cos(\theta), -r\sin(\theta) \right)  \quad \mathbf{x}_{r\theta} = \left(0, -\sin(\theta), \cos(\theta)\right)
\end{align}
Here $'$ denote the $r$-derivative.
Then
\begin{align}
I=ds^2 =(\mathbf{x}_r du + \mathbf{x}_\theta dv) \cdot(\mathbf{x}_r dr + \mathbf{x}_\theta d\theta) = Edr^2 + 2F drd\theta+ G d\theta^2
\end{align}
with $E = 1 +(f')^2$, $F=0$, $G=r^2$. Note
\begin{align}
&\mathbf{x}_r \times \mathbf{x}_\theta = \left(r, -f'(r) r \cos(\theta), -f'(r) r \sin(\theta) \right) \\
&||\mathbf{x}_r \times \mathbf{x}_\theta|| = \sqrt{EG-F^2} = r\sqrt{1 + (f')^2} \\
& U=\frac{\mathbf{x}_r \times \mathbf{x}_\theta}{||\mathbf{x}_r \times \mathbf{x}_\theta||}= \frac{\left(1, -f' \cos(\theta), -f' \sin(\theta) \right)}{\sqrt{1 + (f')^2}}
\end{align}
and
\begin{align}
&e= \mathbf{x}_{rr} \cdot U=\frac{f''}{\sqrt{1 + (f')^2}},  \qquad  f=\mathbf{x}_{r\theta}\cdot U=0 , \qquad g=\mathbf{x}_{\theta\theta} \cdot U=\frac{f'r}{\sqrt{1 + (f')^2}} \\
&\kappa_\mu = \frac{e}{E} =\frac{f''}{\sqrt{\left(1 +(f')^2 \right)^3}}, \quad \kappa_{\pi} = \frac{g}{G}=\frac{f'}{r\sqrt{1+(f')^2}}
\end{align}
Thus we obtain Gauss Curvature
\begin{align}
K = \kappa_\mu \kappa_\pi =\frac{f'f''}{r\left( 1 +(f')^2 \right)^2}\stackrel{f =r^2 }=\frac{4}{\left( 1 +4r^2 \right)^2} >0
\end{align}
