Let $f\in C^1(\mathbb{R}^3, \mathbb{R}^2)$, $f(x,y,z)=(x^2+y^2,y^2+z^2)$ and let $M:=f^{-1}(\{(1,4)\})$, $p=(1,0,2)\in M$.
So I tried sketching $M$ - and I got:
The question is:
Choosing an neighbourhood $V\subset \mathbb{R}^3$ of $p$, find an open intervall $I \subset \mathbb{R}$ and a $C^1$ curve such that $\gamma > (I)=M\cap V$.
Here's another sketch:
I tried finding the black curve by using:
$x^2+y^2=1$ and $(x-1)^2+y^2+(z-2)^2=1$, so $x^2+y^2=(x-1)^2+y^2+(z-2)^2$
solving for $z=2\pm \sqrt{2x-1}$ and $y=\sqrt{1-x^2}$, so $\gamma (t)=(t, \sqrt{1-t^2}, 2\pm \sqrt{2t-1})$... is this correct?
It looks to me that you have the wrong pictures.
We have $$\begin{cases}x^2+y^2=1 \\ y^2+z^2=4\end{cases}$$ This is the intersection of 2 cylinders.
It does not involve a circle as in your first picture, nor a sphere as in your second picture.
At point p=(1,0,2) both the x-coordinate and the z-coordinate take a maximum value.
Let us therefore pick $y=t$ as parameter.
Then $\gamma:(-1,1)\to \mathbb R^3$ with $$\gamma(t)=(\sqrt{1-t^2},\ t,\, \sqrt{4-t^2})$$ does the job.
Alternatively, we can pick polar coordinates for x and y, and subsequently calculate z, so that $\gamma:(-\pi,\pi)\to \mathbb R^3$ with: $$\gamma(\phi)=(\cos\phi, \sin\phi, \sqrt{4-\sin^2\phi})$$