Let $M:=\{(x_1,x_2,x_3)\in\mathbb{R}^3\colon x_1^2+x_2^2-2x_3^2=x_1+x_2+x_3-1=0\}$ be a given submanifold of $\mathbb{R}^3$. I want to compute $T_p(M)$ at $p=(1,1,-1)$.
My definition of the tangential is the following:
Let $Y$ be open in $\mathbb{R}^l$ and $f\in C^1(X,Y)$. Then the linear map
$T_pf\colon T_pX\longrightarrow T_{f(p)}Y, \ (p,v)\mapsto (f(p),\partial f(p)v)$
is called the tangential of $f$ at the point $p$.
Furthermore, i have that for a $(n-l)$-dimensional submanifold $M:=f^{-1}(c)$ where c is a regular value, that $T_pM=\ker(T_pf)=(T_pf)^{-1}(c)$.
What I could do is the following:
I set $\Phi\colon \mathbb{R}^3\longrightarrow \mathbb{R},\ (x_1,x_2,x_3)\mapsto ( x_1^2+x_2^2-2x_3^2,x_1+x_2+x_3-1)$
and already know that $\Phi^{-1}(0)$ is the submanifold in question.
I now need the matrix for $T_pf$ and I struggle to construct it. The Jakobi Matrix of $\Phi$ is no problem. Do I just have to compute $f(p),\partial f(p)$ and put it in the form $(f(p),\partial f(p)v)$ like mentioned above?
Let $f=x_1^2+x_2^2-x_3^2$ and $g=x_1+x_2+x_3-1$.
Being the intersection of the two 2-D Manifolds F given by $f=0$ and G, given by $g=0$, the intersection (call it C) is a 1-D submanifold of the both of them. The point $p$ lies in C and we want to know a tangent vector to C at $p$ and that would define the tangent space entirely (since it is a 1-D space, knowing any vector in it fully specifies it). Note that because C lies in both F and G, a tangent vector to C at $p$ must be perpendicular to normals to both F and G (this tangent vector belongs in both tangent spaces $T_pF$ and $T_pG$). In other words, the cross product $$\nabla f \times \nabla g$$ must be in the tangent space of C at $p$.
The computation of the vector is trivial. So I’ll leave it to you to take it from here.
Second approach
To use your defintions of the tangent space (kernel of the differential map). The tangent space of F at P is $(Df)^{-1}(0)$ and likewise for G. The tangent space of C is $ (Df)^{-1}(0) \cap (Dg)^{-1}(0) $. Noting that for $f:\mathbb{R}^3\to\mathbb{R}$, at the point $p$, $Df$ is given by the matrix $[2,2,4]$ and $Dg$ by the Jacobian $[1,1,1]$. So, we are looking for tangent vectors of the form $\vec{t}=(u,v,w)$ that satisfy the matrix equation.
$$\left[\begin{array}{cc}1 &1&1\\2&2&4\end{array}\right] \left[\begin{array}{cc}u \\v\\w\end{array}\right]= \left[\begin{array}{cc}0\\0\end{array}\right] $$
That is, vectors proportional to $\vec{t}=(-1,1,0)$ make up the tangent space at $p$. You can write down the coordinates of the tangent line if you want as $\vec{p}+s\vec{t}$ where you let the parameter $s$ take all values on the real line.