Determine the tangent planes of $x^2+z^2=1$ at the points $(x, 0, z)$ and show that they are all parallel to the $y-$axis.
Attempt:
Let $f(x, y, z)=x^2+z^2-1$. Then we have that
$$\nabla f(x, y, z)=(2x, 0, 2z)$$
Now, the normal vector of the tangential plane at point $(x, 0, z)$ is
$$\nabla f(x, 0, z)=(2x, 0, 2z)=2(x, 0, z)$$
Since
$$(x, 0, z)\cdot (0,1,0)=0$$
it follows that the vector $2(x, 0, y)$ is perpendicular to the plane $y-$axis, which implies that the tangential plane is parallel to it.
This is another way to write the tangent space to the given subset in a point $p$.
Let $M:=\{(x,y,z)\in\mathbb R^3|x^2+z^2=1\}\subset \mathbb R^3$ be a subspace of the real space.
Define the map $\mathcal C^{\infty}(\mathbb R^3)\ni f:\mathbb R^3\to\mathbb R$ s.t. $f(x,y,z)=x^2+z^2$.
The differential of $f$ in a point $p\in M$ is a function of form $df_p:T_p\mathbb R^3\to T_{f(p)}\mathbb R\cong\mathbb R$ and it is represented by $\nabla f=(2x,0,2z)$, which is singular iff $x,y,z=0.$
We can easily observe that $0\notin f^{-1}(1)=M\implies$ $f$ is a defining map for $M$, in particular its differential on $M$ is surjective.
We can now identify $\text{Ker}(df_p)$ with $T_pM$ for a generic point $p=(a,0,b),$ with $a,b\in\mathbb R.$
$\textbf v=(v_1,v_2,v_3)^t \in T_pM\iff df_p(\textbf v)=0\iff av_1+bv_3=0$.