Let $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ be a smooth function.
Let $G:=\{(x, y, f(x, y)) : x,y \in \mathbb{R}^2\}$ be its graph.
Find a basis for $T_pG$ for a $p(x,y,z) \in G$.
What I did:
I am trying first to determin $T_pG$.
f is smooth so $df: T\mathbb{R}^2 \cong \mathbb{R}^2 \rightarrow T\mathbb{R} \cong \mathbb{R}$.
And since $T_pG=Im(v \rightarrow (v,df_0(v))$ for $v\in T_pG$, can we say that $T_pG=Im(v \rightarrow (v,df_0(v))$ for $v\in\mathbb{R}^2$?
and if yes, how can we go from there? I think I miss something fundamental in understanding manifiolds defined by graph.
Thank you!
** $\textbf{EDIT} $ **
Following up on Andeas's answer, let's consider the map $g:(x,y) \rightarrow (x,y,f(x,y))$
g is clearly injective, smooth with an inverse $h:(x,y,z) \rightarrow (x,y)$ that is smooth so g is an homeomorphism over its image $g(\mathbb{R}^2)$.
In addition, g is smooth and $dg: T\mathbb{R}^2 \cong \mathbb{R}^2 \rightarrow TG$. dg is clearly injective so g is an immersion.
We may conclude that g is a parametrization of G and that a basis of $TG$ is the image of $\{(1,0),(0,1)\}$ by g, i.e.:
$\{(0,1,f(0,1)) (1,0,f(1,0))\}$.
Hint: I think it will be easier to approach the problem by observing that you can nicely parametrize the graph by the map $\mathbb R^2\to G$ defined by $(x,y)\mapsto (x,y,f(x,y))$. Then you can look at the image of the standard basis of $\mathbb R^2$ under the derivative of this parametrization.
Edit (based on the edit of the question):The first part of what you are writing is fine.But then you should use $dg$ rather than $g$ in order to identify the tangent spaces of $G$. The resulting basis for the tangent space $T_{(x,y,f(x,y))}G$ is $\{(1,0,\frac{\partial f}{\partial x}(x,y)), (0,1,\frac{\partial f}{\partial y}(x,y))\}$. In a basis free way, you can describe the tangent space as $\{(v,df(x,y)(v)):v\in\mathbb R^2\}$. You can also view this as the fact that the tangent space to the graph of $f$ in a point $(x,y)$ is the graph of $df(x,y)$.