Taylor Polynomial of $\arctan$ of given Degree and Error 
Replace the following function by its taylor polynomial of the given
  grade, and approximate the error in the given interval:
$$f(x) = \arctan(x) \textrm{ by } T_3(f,x,0) \textrm{ in } |x| \le\frac{1}{10}$$

My solution and thoughts
We only need the first three derivatives and the fourth one for the error
$$
f'(x) = \frac{1}{1+x^2} \\
f''(x) = \frac{2x}{(x^2+1)^2} \\
f'''(x) = \frac{6x^2-2}{(x^2+1)^3} \\
f^{(iv)}(x) = \frac{24x(x^2-1)}{(x^2+1)^4}
$$
And by definition we know that
$$
T_3(f,x,0) = \sum\limits_{k=0}^3 \frac{f^{(k)}(0)}{k!}x^k
$$
we get
$$
T_3(f,x,0) = x - \frac{x^3}{3}
$$
We know that 
$$R_3(x) = \frac{f^{(iv)}(c)}{4!}x^4, |x| \le \frac{1}{10}$$
by plugging in the maximal value of $x = c = \frac{1}{10}$ we get for any $c$:
$$
\left|f^{(iv)}(c)\right| = \left|\frac{24c(c^2-1)}{(c^2+1)^4}\right| \le \\
\le \frac{24\cdot 99 \cdot 100^4}{1000\cdot 101^4}
$$
$$
\Rightarrow \left|R_3(x)\right| \le \frac{24\cdot 99 \cdot 100^4}{1000\cdot 101^4 \cdot 4!}\cdot\left|x^4\right| \Leftrightarrow \\
\Leftrightarrow \left|R_3(x)\right| \le \frac{10}{101^4}
$$
Is this solution correct? How can I make it more formally right? (I know I lack some mathematical formalism).

Your work looks good to me, except for a mistake in the last step (probably a typo).  Other than that, it's pretty much exactly the way I would hope one of my students would solve this problem.
The mistake is that you lost the $99$ in the final step, so that at the end you should have $$
\left|R_3(x)\right| \le \frac{99 \cdot 10}{101^4}.$$
(Also, the final implication is $\Rightarrow$, not $\Leftrightarrow$.  Since you're substituting a value for $x$ here the two inequalities are not equivalent.)

As a side note, there's an easier way to calculate the Taylor series about $0$ for $f(x) = \arctan x$.
Since $$\frac{d}{dx} \arctan x = \frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 \pm \ldots,$$ via the geometric series formula when $|x| < 1$, 
$$\arctan x = C + x - \frac{1}{3} x^3 + \frac{1}{5}x^5 - \frac{1}{7} x^7 \pm \ldots.$$
Since $\arctan 0 = 0$, we have  $C = 0$.  This gives you $$T_3(f,x,0) = x - \frac{1}{3}x^3$$ rather painlessly.
You do still need to find the fourth derivative of $\arctan x$ to use the remainder formula for Taylor polynomials, however.
