Given $A^2$ where A is matrix, how find A?

Problem is simple.

Given $$A^2=\begin{bmatrix}13 & 9 &-9 \\ 0 & 4 & 0 \\ 12 & 12 & -8 \end{bmatrix}$$ How find $A$?

I think a method using eigenvalues and I find them. But I can't find an actual $A$. Is it right to use eigenvalues?

I bet that when you say eigenvalues, you mean that you might diagonalize the matrix $A^2$ by expressing it as $A^2 = VDV^{-1}$ for some matrix $V$ and diagonal matrix $D$, where the entries in $D$ are precisely the eigenvalues of $A^2$.

Once you've done this, it is very easy to find a square root of $D$, since it is a diagonal matrix. For example, a square root of the matrix $\begin{pmatrix} 4 & 0 \\ 0 & 9 \end{pmatrix}$ might be $\begin{pmatrix} 2 & 0 \\ 0 & 3 \end{pmatrix}$.

Does this trail of breadcrumbs lead you to the answer?

Actually you don't need eigenvectors, just eigenvalues (there is some complication in the case of eigenvalue $0$, but fortunately this example is nonsingular). Find a polynomial $g(t)$ with the following property: for each eigenvalue $\lambda$, $g(\lambda) = \sqrt{\lambda}$ and, if $\lambda$ has multiplicity $k > 1$ in the minimal polynomial of $A^2$, the first $k-1$ derivatives of $g$ and $\sqrt{}$ agree at $\lambda$. Then take $A = g(A^2)$. This will work even when there does not exist a basis of eigenvectors, as long as $0$ is at most a simple zero of the minimal polynomial.