Fix $\varepsilon > 0$ and $a$.
From the definition of differentiation we have
$$
\left|\frac{f(x)-f(a)}{x-a}-f'(a)\right| < \varepsilon
$$
for an appropriately chosen $\delta > 0$.
Multiply both sides by $|x - a|$ to get:
$$
\left|f(x) - f(a) - (x - a)f'(a)\right| < |x - a| \varepsilon
$$
Using $\left||x|-|y|\right| \le |x - y|$ we have:
$$
\left|f(x) - f(a)\right| - |x - a| \cdot \left|f'(a)\right| < |x - a| \varepsilon
$$
Rearrange to get:
$$
\left|f(x) - f(a)\right| < (\left|f'(a)\right| + \varepsilon) \cdot |x - a|
$$
Since $f'(a)$ and $\varepsilon$ are both fixed, you can make $|f(x) - f(a)|$ as small as you want by making $|x - a|$ smaller and smaller. Thus, the function is continuous at $a$.
To prove this formally, pick any $\hat{\varepsilon}$ (different from $\varepsilon$ fixed at the beginning and used with the differentiation definition). Pick $\hat{\delta} = \min\left(\delta, \frac{\hat{\varepsilon}}{\left|f'(a)\right| + \varepsilon}\right)$. Clearly:
$$
|x - a| < \hat{\delta} \Rightarrow \left|f(x) - f(a)\right| < \hat{\varepsilon}
$$
If you can show that
$$
\forall\epsilon >0 \ \exists\delta>0 \ s.t. 0 < |x-c| < \delta \implies
\delta = \frac{\epsilon}{|x+3|},
$$
then you might be able to use $\delta = \frac{\epsilon}{|x+3|}$ in your proof.
Can you prove the implication?
The piece above is a broad hint about where you are going wrong.
Here is a more explicit explanation:
You can think of the definition of a limit as a game you play against a perfectly clever opponent.
Your opponent moves first by selecting $\epsilon.$
They can choose any positive number whatsoever, tiny, huge, or in between.
Next, it's your turn; you get to select $\delta.$
You can choose any $\delta$ you like as long as it's positive.
Now it's your opponent's move again.
They get to choose $x,$
but this time the choice is restricted to a punctured neighborhood of $c$ that you defined when you chose $\delta$;
namely, they have to choose $x$ such that $0 < \lvert x - c \rvert < \delta.$
Now if $\lvert f(x) - L \rvert \geq \epsilon,$ you lose.
But to do an actual proof, you need to describe a perfect strategy so that your opponent can never defeat you.
In effect, your formula for $\delta$ is a program for a limit-game-playing robot that plays your part of the game.
Making your task even more difficult, your opponent will be able to know the program before the game starts.
So if there is any flaw in the program, your opponent can exploit it and win.
Flaws include values of $\delta$ that let the opponent set $x$ so that
$\lvert f(x) - L \rvert \geq \epsilon,$
as well as cases where your rule doesn't produce a definite answer.
So let's say your "program" for the robot is
$$ \delta = \frac{\epsilon}{|x+3|}.$$
Your opponent chooses $\epsilon = 10^{-100}.$
Now it's your move. What does your robot choose for $\delta$?
Your rule for $\delta$ requires the robot to know $x,$ which is a number your opponent has not selected yet. So your robot has no way to compute $\delta.$
Poor robot. You lose.
Best Answer
Hint: $$\left|\frac{f(x)-f(c)}{x-c}-2c\right|=\left|\frac{x^2-c^2}{x-c}-2c\right|=\left|\frac{(x+c)(x-c)}{x-c}-2c\right|=\left|x-c\right|$$