You want to solve the equation $(I+X)^2=I+Y$ for the $(n\times n)$-matrix $X$ when $Y$ is a given matrix near $0$. This amounts to
$$Z(X):=2X+X^2\ =\ Y\ ,\qquad (*)$$
an equation in the $n^2$ unknowns $X_{ik}$. When $Y=0$ then $X=0$ is a solution of this equation. According to the assumptions of the implicit function theorem we have to check whether the $(n^2\times n^2)$-matrix
$$J:=\Bigl({\partial(Z_{11},Z_{12},\ldots,Z_{nn})\over
\partial(X_{11},X_{12},\ldots,X_{nn})}\Bigl)_{X=0}$$
is regular. As $Z_{ik}=2 X_{ik} +$ terms of degree $2$ in the $X_{ik}$, the matrix $J$ is $2$ times the $n^2\times n^2$ identity matrix; so it is certainly regular. It follows that the equation $(*)$ has a solution $X=F(Y)$ such that $F(0)=0$ and $F(\cdot)$ is a differentiable matrix valued function of (the matrix elements of) $Y$.
By the way, the square root $I+X$ of $I+Y$ for $\|Y\|<1$ is given by the binomial series
$$\sum_{k=0}^\infty\ {1/2 \choose k}\ Y^k\ .$$
Just use the definition of cosh:
$$ cosh(z) := \frac{e^{z}+e^{-z}}{2} > 0 \Rightarrow 2cosh(y)+3cosh(3x) > 0$$
The image of the exponential function is always greater zero. This contradicts:
$$\exists x,y \in \mathbb{R}:2cosh(y)=-3cosh(3x) \Leftrightarrow 2cosh(y)+3cosh(3x) =0 $$
Best Answer
No. What you should concern about is not the invertibility/non-invertibility of the matrix $AH+HA$, but the invertibility/non-invertibility of the linear map $H\mapsto AH+HA$. In other words, the square map $A\mapsto A^2$ has a local inverse at $A=\begin{pmatrix}1&2\\-2&1\end{pmatrix}$ if the equation $AH+HA=0$ has only the trivial solution $H=0$.
Edit: Note that what the inverse function theorem gives is a sufficient condition rather than a necessary condition for the existence of a local inverse. So, if the derivative mapping $H\mapsto AH+HA$ turns out to be singular, all you can say is merely that the inverse function theorem doesn't apply. To prove that the square map has no local inverse at $A$, you must use other methods. For example, you may find two sequences of matrices $\{B_n\}$ and $\{C_n\}$ such that $\lim_{n\to\infty}B_n=\lim_{n\to\infty}C_n=A$, but $B_n\not=C_n$ and $B_n^2=C_n^2$ for every $n$. Alternatively, you may find a sequence of matrices $\{S_n\}$ such that $\lim_{n\to\infty}S_n=A^2$, but $S_n$ has no (real) square root. To this end you may consider $S_n=\begin{pmatrix}-3&1/n\\0&-3\end{pmatrix}$.