I have to find a sufficient condition for $f(0)$ for having $f(x)>0$, if $f:[0,3]\to\mathbb{R}$ continuous and differentiable, and $f'(x)>-1$ for $x\in(0,3)$.
Since $f$ is differentiable in (0,3) if I apply MVT I have that $\exists c\in(0,x)$ with $0<x<3$ such that $f'(c)=\frac{1}{x}(f(x)-f(0))$. So: $$f(x)-f(0)>xf'(c)>-x\Longrightarrow f(x)>f(0)-x$$ So a sufficient condition for having $f(x)>0$ is: $f(0)>x>3\Longrightarrow f(0)>0$
Am I right?
Just the last part of your reasoning is incorrect. We have $f(x)>f(0)-x$, and we want $f(x)>0$ for all $x\in[0,3]$.
A sufficient condition for $f(x)>0$ would be $f(0)-x>0$ (since we would then have $f(x)>f(0)-x>0$), or equivalently $f(0)>x$.
Since $x\in[0,3]$, thus $f(0)>x$ for all $x$ requires $f(0)>3$.