Proving that if $Y$ is a real matrix, $e^Y$ can't have negative eigenvalues. Prove: for the matrix $ A=\pmatrix {
    \begin{matrix}
    -1 & 0  \\
    6 & 3  
    \end{matrix}}$
 there doesn't exist a real matrix $X$ such that $e^X=A$.
What I tried: lets assume the existence of such matrix $X$. Diagonalizing $A$ gives $ J=\pmatrix {
    \begin{matrix}
    3 & 0  \\
    0 & -1  
    \end{matrix}}$
That means for some $P$ we have $P^{-1}AP=J  \; \Rightarrow \; P^{-1}e^XP=J$.
Denoting $Y=P^{-1}XP\; $ we get $e^Y=P^{-1}e^XP \; \Rightarrow e^Y= \pmatrix {
    \begin{matrix}
    3 & 0  \\
    0 & -1  
    \end{matrix}}$ 
where $Y$ is a real matrix.
I think I'm close to a condtradiction. I want to say that because $Y$ is real, $e^Y$ can't have negative eigenvalues. Is that correct? how can I show that?  
 A: Note that $\det A<0$. If $A=\exp Y $, then$$\det A=e^{\operatorname{tr}Y}>0.$$
A: It's not true that if $Y$ is real, $e^Y$ can't have negative eigenvalues.
Try
$$ Y = \pmatrix{0 & \pi\cr -\pi & 0\cr},\ e^Y = \pmatrix{-1 & 0\cr 0 & -1}$$
This is also a counterexample to Surb's comment.
A: I think it's better if you diagonalize $X$. But then, maybe $X$ is not diagonalizable. 
Now, note the following. If $Xv=\lambda v$, then $X^jv=\lambda^j\,v$ for all integer $j$, and so 
$$
e^Xv=\sum_{j=0}^\infty \frac1{j!}\,X^jv
=\sum_{j=0}^\infty\frac{\lambda^j}{j!}\,v=e^\lambda\,v.
$$
That is, if $\lambda$ is an eigenvalue for $X$, then $e^\lambda$ is an eigenvalue for $e^X$. In your particular case, since $A$ has two distinct eigenvalues, the above requires $X$ to have two distinct eigenvalues. Being $2\times2$, this means that $X$ is diagonalizable. 
So $X=QHQ^{-1}$, with $H$ diagonal, and 
$$
A=e^X=Qe^HQ^{-1}=Q\,\begin{bmatrix} e^{h_1}&0\\0&e^{h_2}\end{bmatrix} \,Q^{-1}.
$$
In particular the eigenvalues of $A$ are exponentials of real numbers, so they cannot be negative (nor zero). 
