Existence of differentiable matrix maps $M(3,\mathbb{R}) \rightarrow M(3,\mathbb{R})$

We use the implicit function theorem, that is a well-known method.

i) for$f$. Let $p:X\in M_3\mapsto X^2$ and $U=diag(-1,1,1)$; then $p(U)=I_3$; the derivative of $p$ is

$Dp_X:H\in M_3\mapsto XH+HX$ and $Dp_U(H)=UH+HU$ is a sum of functions that commute.

$p$ has a local $C^{1}$ inverse from a neighborhood of $I_3$ to a neighborhood of $U$ IFF $Dp_U$ is invertible. Let $spectrum(U)=(\lambda_i)_i$.

According to

https://en.wikipedia.org/wiki/Kronecker_product#Abstract_properties

$spectrum(Dp_U)=\{\lambda_i+\lambda_j;i,j\}=\{-2,0,0,0,0,2,2,2,2\}$ and, therefore $p$ has no $C^1$ local inverse.

ii) for $g$ (in the same way). Let $q:X\in M_3\mapsto X^3$ and $V=\begin{pmatrix}0&1&0\\0&0&1\\1&0&0\end{pmatrix}$; then $q(V)=I_3$; the derivative of $q$ is

$Dq_X:H\in M_3\mapsto HX^2+XHX+X^2H$ and $Dq_V(H)=HV^2+VHV+V^2H$ is a sum of functions that commute.

$spectrum(V)=spectrum(V^2)=(\mu_i)_i=\{1,u,u^2\}$ where $u=e^{2i\pi/3}$.

Then $spectrum(Dq_V)=\{\mu_i^2+\mu_i\mu_j+\mu_j^2;i,j\}$.

With $\mu_i=1,\mu_j=u$, we obtain (at least) one zero eigenvalue and, therefore, $q$ admits no local $C^1$ inverse.

EDIT. Answer to the OP and Sally G.

If you don't know the theory of Kronecker products, then, no matter, it suffices to display elements of $\ker(Dp_U)$ and of $\ker(Dq_V)$. For example

$H=\begin{pmatrix}0&0&0\\1&0&0\\0&0&0\end{pmatrix}\in\ker(Dp_U)$ and $H=diag(1,0,-1)\in\ker(Dq_V)$.