For $f: \mathbb{R} \rightarrow \mathbb{R}$ it holds that if $f \circ \zeta_1$ is total differentiable in $(0,0)$ then $f$ is differentiable in $0$. Suppose $\zeta: \mathbb{R}^2 \rightarrow \mathbb{R}^2$ is a $C^1$ function that is total differentiable with $\zeta (0,0) = (0,0)$ and where $(d\zeta)(0,0)$ is invertible. Write $\zeta_1$ as the first component function.
Show that for any function $f: \mathbb{R} \rightarrow \mathbb{R}$ it holds that if $f \circ \zeta_1$ is total differentiable in $(0,0)$ then $f$ is differentiable in $0$.
I'm assuming the trick has to do with the invertibility of $\zeta$, but I'm stuck on the proof. Below is what I already found.
Because $\zeta$ is $C^1$ and $(d\zeta)(0,0)$ is invertible we can use the inverse function theorem to get in the neighborhood of $(0,0)$ a locally total differentiable inverse of $\zeta$, say $g$, such that $(dg)(0,0) = [(d\zeta)(0,0)]^{-1}$ (after some simplification). I'm stuck here, any tips?
Edit: I constructed a function $h: \mathbb{R}^2 \rightarrow \mathbb{R}^2: (x,y) \mapsto (f(x),f(x))$. Note that $h \circ \zeta \circ g$ is total differentiable because $h \circ \zeta$ is in both components and $g$ is by the inverse function theorem. But now $h \circ \zeta \circ g = h$ and thus $h$ is total differentiable and hence $f$ is differentiable in $0$.
Note that this is just a sketch of my proof.
 A: Your edit seems fine. The rigorous proof (of what I suggested in the comments) is as follows: let $\iota_1:\Bbb{R}\to\Bbb{R}^2$, $\iota_1(x)=(x,0)$ and $\pi_1:\Bbb{R}^2\to\Bbb{R}$, $\pi_1(x,y)=x$ be the canonical injections and projections. Then, $\zeta_1$ is by definition equal to $\pi_1\circ \zeta$ "the first component of $\zeta$". Now, note that
\begin{align}
(f\circ\zeta_1)\circ(\zeta^{-1})\circ(\iota_1)&=f\circ(\pi_1\circ\zeta)\circ\zeta^{-1}\circ\iota_1\\
&=f\circ\pi_1\circ\iota_1\\
&-f\circ\text{id}_{\Bbb{R}}\\
&=f
\end{align}
($\pi_1\circ\iota_1=\text{id}_{\Bbb{R}}$ is clear since for any $x\in\Bbb{R}$, $\pi_1(\iota_1(x)):=\pi_1(x,0)=x$). Hence, we have expressed $f$ on the right as a triple composition of functions each of which is differentiable at the origin ($f\circ \zeta_1$ is differentiable by hypothesis; $\zeta^{-1}$ is differentiable at origin since $\zeta(0)=0$, and by the inverse function theorem as you mentioned, and finally $\iota_1$ is a linear transformation, so clearly is differentiable everywhere, in particular at the origin).
In short, the above is simply expressing that for all $x\in\Bbb{R}$, $f(x)=(f\circ \zeta_1\circ \zeta^{-1})(x,0)$.
