For a function to begin to qualify as differentiable, it would need to be continuous, and to that end you would require that [tex]a[/tex] is such that
[tex]\displaystyle\lim_{x\to0^-}g(x)=\lim_{x\to0^+}g(x)\iff\lim_{x\to0}ax=\lim_{x\to0}x^2-3x[/tex]
Obviously, both limits are 0, so [tex]g[/tex] is indeed continuous at [tex]x=0[/tex].
Now, for [tex]g[/tex] to be differentiable everywhere, its derivative [tex]g'[/tex] must be continuous over its domain. So take the derivative, noting that we can't really say anything about the endpoints of the given intervals:
[tex]g'(x)=\begin{cases}a&\text{for }x<0\\2x-3&\text{for }x>0\end{cases}[/tex]
and at this time, we don't know what's going on at [tex]x=0[/tex], so we omit that case. We want [tex]g'[/tex] to be continuous, so we require that
[tex]\displaystyle\lim_{x\to0^-}g'(x)=\lim_{x\to0^+}g'(x)\iff\lim_{x\to0}a=\lim_{x\to0}2x-3[/tex]
from which it follows that [tex]a=-3[/tex].
