wrongly asked question about precalculus? Im seeing the following question in a precalc textbook: Suppose $f$ is a function whose domain is $[-5,5]$ and $f(x) = \frac{x}{x+3}$ for every $x$ in $[0,5]$. Suppose $f$ is an odd function. Evaluate $f(-3)$.

Isnt this a bad formulated problem?? I mean, $f(-3)$ is not even defined, but since $f$ is odd, then $f(-3) = - f(3) = - \frac{1}{2}$. 

Or, am I just misinterpreting the question?
It is not poorly formulated but may seem a bit tricky. This essentially is a piecewise function as

$f(x)=\frac{x}{x+3}: \quad x\in [0,5]$

$f(x)=-\frac{|x|}{|x|+3}: \quad x\in [-5,0)\quad\!\!$ because the function is odd. 

Thus, $f(-3)=-\frac{1}{2}$ as you have mentioned.
To give a direct answer using the properties that you have likely already covered. We are given that $f$ is an odd function, meaning that it satisfies the equation

$f(-x)=-f(x)$

Hence to find $f(-3)$ we let $x=3$ and apply the above formula, yielding

$f(-3)=-f(3)$

I assume you know how to find $f(3)$ using the given formula, which defines $f$ for $x$ in $[0,5]$. (Note the negative sign on $f(3)$ in the above formula, and be sure to include it in your final answer!)