Differentiating $x+2x^2\sin(1/x)$ near $0$, discontinuity of the derivative I have this function
$$
\begin{array}{l}
f:\mathbb{R}\rightarrow\mathbb{R}\\
x\rightarrow\left\{\begin{array}{ll}
x+2x^2\sin\left(\frac{1}{x}\right)&x\neq 0\\
0&x=0
\end{array}\right.
\end{array}
$$
I need to calculate its derivative in $x=0$. I'm not sure how can I compute it at zero, since the limit of its derivative goes to 1 when x tends to zero but at zero I thought $f'(0)$ must be zero (therefore its derivative is not continuous at the origin). If is not zero then it is invertible around the origin via the Inverse Function Theorem, right?
 A: The derivative is discontinuous at the origin, but not for the reasons you stated.
Calculate the derivative at $x=0$ using the limit definition, $f'(0)=\lim\limits_{h\to 0}\dfrac{f(h)-f(0)}{h}$.  You will find that it is $1$, not $0$.
Calculate the derivative at $x\neq 0$ using the ordinary shortcuts from calculus.  You will have a general expression for $f'(x)$ when $x\neq 0$, and you will find that $\lim\limits_{x\to 0}f'(x)$ does not exist.

I seem to have neglected the question about invertibility.  A continuous function on an interval in $\mathbb R$ is invertible if and only if it is strictly monotone (a consequence of the intermediate value theorem).  A function with everywhere existing nonzero derivative in an interval is strictly monotone (a consequence of the mean value theorem). In particular, if a function has a continuous derivative and nonzero derivative at zero, then the function is invertible in a neighborhood of the origin, and the inverse function theorem even applies.  
For your function, although $f'(0)$ exists and is nonzero, discontinuity of $f'$ means the inverse function theorem doesn't apply, and more work is required to determine whether $f$ is invertible.  A necessary condition for a differentiable function to be monotone on an interval is that the derivative doesn't change signs on the interval.  For your function, $f'(0)=1>0$, but evaluating at $x_n = \dfrac{1}{2n\pi}$ gives $f'(x_n)=-1$ with $x_n\to 0$ as $n\to \infty$.  Hence, $f$ is not monotone in any interval containing $0$.
A: Use the definition of the derivative to evaluate it at zero:
$$f'(0) = \lim_{x \to 0} \frac{f(x) - f(0)}{x - 0}.$$
In this case we have
\begin{align}
\lim_{x \to 0} \frac{f(x) - f(0)}{x - 0} & = \lim_{x \to 0} \frac{x + 2x^2 \sin(1/x)}{x} \\
& = \lim_{x \to 0} (1 + 2x \sin(1/x)) \\
& = 1 + 0 = 1,
\end{align}
where we use the squeeze theorem to conclude that
$$\lim_{x \to 0} 2x \sin(1/x) = 0.$$
