Let $A\subset \mathbb{R}^n$ an open set, and $g:A\to \mathbb{R}$ continously differentiable such that $g'(x)\not=0 $ for $x\in A$. If $M = g^{-1}(\{0\})\not=\emptyset$, then I want to show that the tangent space in $x\in M$ is
$$\{v_x\in \mathbb{R}^n_x: v \cdot \nabla g(x) = 0 \}$$
Note: We have defined as the tangent space the image $f_{*}(\mathbb{R}^n_x)$ where $f_{*}$ is defined as follows, Let $M^{k}$ be a differentiable manifold in $\mathbb{R}^n$ and $f: U \to V$ a coordinate system around $p=f(x)$ so we define the linear transformation $f_{*}:\mathbb{R}^k_x \to \mathbb{R}^n_x$ as
$$f_{*}(v_x)= Df(x)(v)_{f(x)}$$
My attempt:
This result for me is clear, but the problem Is how to write this formally, then I want to say that since $M$ is a differentiable manifold we have a coordinate system $f$ , now, what I want to have is the Jacobian matrix and then compute its image, the thing is that I don't know who is $f$ explicitly (I have tried to see M as level curves, but I don't know how that thing may help me to explain the things better). Can you help to fix this an give to all of this order? please
I deal with the more general case where the manifold $M$ is (locally) defined by $q$ equations $g_i(x)=0$, $1\leq i\leq n$, $g_i\in\mathcal{C}^k\left(A;\mathbb{R}\right)$ where $k$ is the smoothness of $M$, AND the $Dg_i(x)$ are linearly independant.
Because the gradient $\nabla g_i(x)$ is defined as follow : $$\left\langle \nabla g_i(x),v\right\rangle=Dg_i(x)[v]$$ for all vector $v$ (where $\left\langle .,.\right\rangle$ denotes the inner product of $\mathbb{R}^n$), we must show that $$T_xM=\bigcap_{i=1}^n\ker Dg_i(x)$$ where $T_xM$ is the tangent (vector) space at the point $x$.
If $x\in M$, then $g_i(x)=0$, hence $Dg_i(x)=0$ so that any tangent vector at $x$ is in $\ker Dg_i(x)$, $1\leq i\leq n$.
Consider now $v\in \bigcap_{i=1}^n\ker Dg_i(x)$. By the assumption on the differentials $Dg_i(x)$, the matrix $$\begin{pmatrix}\frac{\partial g_{1}}{\partial x_{1}}(x) & \ldots & \frac{\partial g_{1}}{\partial x_{q}}(x) & \ldots & \frac{\partial g_{1}}{\partial x_{n}}(x)\\ \vdots & & \vdots & & \vdots\\ \frac{\partial g_{q}}{\partial x_{1}}(x) & \ldots & \frac{\partial g_{q}}{\partial x_{q}}(x) & \ldots & \frac{\partial g_{q}}{\partial x_{n}}(x) \end{pmatrix} $$ has a rank equals to $q$. We can permute its columns so that the matrix $\left(\frac{\partial g_{i}}{\partial x_{h}}(x)\right)_{1\leq i,j\leq q}$ is invertible. We then write $$\mathbb{R}^n\simeq\mathbb{R}^q\oplus\mathbb{R}^{n-q}\quad,\quad x=(x',x'')\quad,\quad v=(v',v'')$$ so that if $g=\left(g_1,\ldots,g_q\right):A\longrightarrow\mathbb{R}^q$, then $$0=Dg(x)[v]=Dg(x)_q[v']+Dg(x)_{n-q}[v'']$$ where $Dg(x)_q$ and $Dg(x)_{n-q}$ are the partial differentials of $g$ on $\mathbb{R}^q$ and $\mathbb{R}^{n-q}$, respectively. Hence $$v'=-Dg_{q}(x)^{-1}\left(Dg_{n-q}(x)[v'']\right).$$ We have proven that $g$ as an invertible partial differential : by the implicit functions theorem, we can (locally) write $h(x'')=x'$ where $h$ is the "implicit function", satisfying $Dh(x'')[v'']=v'$. Let us define the curve $$\gamma(t)=(\underbrace{h(x''+tv'')}_{\in\mathbb{R}^q},\underbrace{x''+tv''}_{\in\mathbb{R}^{n-q}})\in\mathbb{R}^n.$$ Then $$\gamma(0)=\left(h(x''),x''\right)=(x',x'')=x,$$ $$\dot{\gamma}(0)=\left(Dh(x'')[v''],v''\right)=(v',v'')=v.$$ We usually define the tangent (vector) space as the set of all such derivatives. But if we use the definition of the question, then each component $f_i(x)$ for $1\leq i\leq q$ can be taken as the $i$-th component of the vector $h(x''+tv'')\in\mathbb{R}^q$ and each component $f_j(x)$ for $q+1\leq j\leq n$ can be taken as the $(j-q)$-th component of the vector $x''+tv''\in\mathbb{R}^{n-q}$ : doing this, we have written the initial vector $v$ as $f_*(v)$ at the point $x$.