I am afraid that the theorem that you quoted will be of no use. Nevertheless what you can do is as follows.
First of all it should be $\lim_{x\to0^{+}}f(x) = 3$ instead of $\lim_{x\to0}f(x) = 3$ as $f'(x)$ exists only for $x>0$ and in that case $f'(0)$ may not exist, for example, consider the function $f(x) = |3x|$.
And if you are assuming that $\lim_{x\to0}f(x) = 3$, then you are assuming implicitly that $f'(x)$ exists for negative values of $x$ sufficiently close to $0$. In that case $f'(0)$ exists and in fact $f'(0) = 3$. Because showing that $f'(0) = 3$ is equivalent to showing that given $\epsilon >0$ there exists $\delta >0$ such that
\begin{equation}
\biggl{|}\frac{f(x)-f(0)}{x} - 3\biggr{|} < \epsilon
\end{equation}
whenever $|x|<\delta$. Since $\lim_{x\to0}f(x) = 3$, it follows that given $\epsilon>0$, there exists a $\delta_1>0$ such that $|f'(c)-3|<\epsilon$ whenever $|c|<\delta_1$, $c \neq 0$ obviously. Now choose $\delta = \delta_1$. Now for any $x$ with $|x|<\delta = \delta_1$, we can apply Mean-Value Theorem to $f$ on the interval $I_x$ whose end points are $x$ and $-x$. Applying Mean-Value Theorem will give us that there exists $b\in I_x$ such that
\begin{equation}
\frac{f(x)-f(0)}{x} = f'(b)
\end{equation}
In particular,
\begin{equation}
\biggl{|}\frac{f(x)-f(0)}{x} - 3\biggr{|} = |f'(b) - 3| < \epsilon
\end{equation}
whence it follows that $f'(0)$ exists and in fact $f'(0) = 3$.
Note: If $\lim_{x\to0^{+}}f(x) = 3$ is the condition then by the same argument what you can actually show is that $f'(0^{+})$ exists and $f'(0^{+})=3$, where $f'(0^{+})$ denotes the right hand derivative of $f$ at $0$. Just apply the Mean-Value Theorem to the interval $[0,x]$ for $x$ sufficiently close to $0$.
